Calculus

Integration of Functions

Integration of Functions Logo

The Fundamental Theorem of Calculus

  • The Fundamental Theorem of Calculus (FTC) shows that differentiation and integration are inverse processes.

    Part \(1\) (FTC1)

    If \(f\) is a continuous function on \(\left[ {a,b} \right],\) then the function \(g\) defined by

    \[{{g\left( x \right)} = \int\limits_a^x {f\left( {t} \right)dt},\;\;}\kern0pt{{a \le x \le b}}\]

    is an antiderivative of \(f\), that is

    \[{g^\prime\left( x \right) = f\left( x \right)\;\;\text{or}\;\;}\kern0pt{\frac{d}{{dx}}\left( {\int\limits_a^x {f\left( t \right)dt} } \right) }={ f\left( x \right).}\]

    If \(f\) happens to be a positive function, then \(g\left( x \right)\) can be interpreted as the area under the graph of \(f\) from \(a\) to \(x.\)

    The Fundamental Theorem of Calculus (Part 1)
    Figure 1.

    The first part of the theorem says that if we first integrate \(f\) and then differentiate the result, we get back to the original function \(f.\)

    Part \(2\) (FTC2)

    The second part of the fundamental theorem tells us how we can calculate a definite integral.

    If \(f\) is a continuous function on \(\left[ {a,b} \right]\) and \(F\) is an antiderivative of \(f,\) that is \(F^\prime = f,\) then

    \[{\int\limits_a^b {f\left( x \right)dx} }= {F\left( b \right) – F\left( a \right)\;\;}\kern0pt{\text{or}\;\;{\int\limits_a^b {{F^\prime\left( x \right)}dx} }= {F\left( b \right) – F\left( a \right)}.}\]

    To evaluate the definite integral of a function \(f\) from \(a\) to \(b,\) we just need to find its antiderivative \(F\) and compute the difference between the values of the antiderivative at \(b\) and \(a.\)

    So the second part of the fundamental theorem says that if we take a function \(F,\) first differentiate it, and then integrate the result, we arrive back at the original function, but in the form \(F\left( b \right) – F\left( a \right).\)

    Thus, the two parts of the fundamental theorem of calculus say that differentiation and integration are inverse processes.

    The Area under a Curve and between Two Curves

    The area under the graph of the function \(f\left( x \right)\) between the vertical lines \(x = a,\) \(x = b\) (Figure \(2\)) is given by the formula

    \[S = \int\limits_a^b {f\left( x \right)dx} = {F\left( b \right) – F\left( a \right).}\]

    Area under a curve
    Figure 2.

    Let \(F\left( x \right)\) and \(G\left( x \right)\) be antiderivatives of functions \(f\left( x \right)\) and \(g\left( x \right),\) respectively.

    If \(f\left( x \right) \ge g\left( x \right)\) on the closed interval \(\left[ {a,b} \right],\) then the area between the curves \(y = f\left( x \right),\) \(y = g\left( x \right)\) and the lines \(x = a,\) \(x = b\) (Figure \(3\)) is given by

    \[
    {S = \int\limits_a^b {\left[ {f\left( x \right) – g\left( x \right)} \right]dx} }
    = {F\left( b \right) – G\left( b \right) }-{ F\left( a \right) + G\left( a \right).}
    \]

    Area between two curves
    Figure 3.

    The Method of Substitution for Definite Integrals

    The definite integral \(\int\limits_a^b {f\left( x \right)dx} \) of the variable \(x\) can be changed into an integral with respect to \(t\) by making the substitution \(x = g\left( t \right):\)

    \[{\int\limits_a^b {f\left( x \right)dx} }={ \int\limits_c^d {f\left( {g\left( t \right)} \right)g’\left( t \right)dt} .}\]

    The new limits of integration for the variable \(t\) are given by the formulas

    \[{c = {g^{ – 1}}\left( a \right),\;\;}\kern-0.3pt{d = {g^{ – 1}}\left( b \right),}\]

    where \({g^{ – 1}}\) is the inverse function to \(g,\) that is \(t = {g^{ – 1}}\left( x \right).\)

    Integration by Parts for Definite Integrals

    In this case the formula for integration by parts looks as follows:

    \[{\int\limits_a^b {udv} }={ \left. {uv} \right|_a^b – \int\limits_a^b {vdu} ,}\]

    where \(\left. {uv} \right|_a^b\) means the difference between the product of functions \(uv\) at \(x = b\) and \(x = a.\)


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Calculate the derivative of the function \(g\left( x \right) = \int\limits_1^x {\sqrt {{t^3} + 4t} dt} \) at \(x = 2.\)

    Example 2

    Calculate the derivative of the function \(g\left( x \right) = \int\limits_{ – \large{\frac{\pi }{2}}\normalsize}^x {\sqrt {{{\sin }^2}t + 2} dt} \) at \(x = \large{\large{\frac{\pi }{6}}\normalsize}.\)

    Example 3

    Find the derivative of the function \(g\left( x \right) = \int\limits_3^{{x^2}} {\large{\frac{{dt}}{t}}\normalsize}.\)

    Example 4

    Find the derivative of the function \(g\left( x \right) = \int\limits_0^{{x^2}} {\sqrt {1 + {t^2}} dt}.\)

    Example 5

    Find the derivative of the function \(g\left( x \right) = \int\limits_1^{{x^3}} {{t^2}dt}.\)

    Example 6

    Find the derivative of the function \(g\left( x \right) = \int\limits_{{x^2}}^{{x^3}} {tdt}.\)

    Example 7

    Calculate the derivative of the function \(g\left( x \right) = \int\limits_{\sqrt x }^x {\left( {{t^2} – t} \right)dt} \) at \(x = 1.\)

    Example 8

    Evaluate the integral \(\int\limits_0^2 {\left( {{x^3} – {x^2}} \right)dx}.\)

    Example 9

    Evaluate the integral \(\int\limits_{ – 1}^1 {\left( {{t^2} + {t^{21}}} \right)dt}.\)

    Example 10

    Calculate the integral \(\int\limits_0^1 {\left( {\sqrt[\large 3\normalsize]{t} – \sqrt t } \right)dt}.\)

    Example 11

    Evaluate the integral \(\int\limits_0^1 {{\large\frac{x}{{{{\left( {3{x^2} – 1} \right)}^4}}}\normalsize} dx}.\)

    Example 12

    Evaluate the integral \(\int\limits_1^e {\left( {t + \large{\frac{1}{t}}\normalsize} \right)dt}.\)

    Example 13

    Evaluate the integral \(\int\limits_0^{\ln 2} {x{e^{ – x}}dx}.\)

    Example 14

    Evaluate the integral \(\int\limits_{ – 1}^1 {\left| {x – \large{\frac{1}{2}}\normalsize} \right|dx}.\)

    Example 15

    Evaluate the integral \(\int\limits_{ – 2}^1 {\left| {{x^2} – 1} \right|dx}.\)

    Example 16

    Find the area bounded by the curves \(y = {x^2}\) and \(y = \sqrt x.\)

    Example 17

    Find the area bounded by the curves \(y = 2x – {x^2}\) and \(x + y = 0.\)

    Example 18

    Find the area of the triangle with vertices at \(\left( {0,0} \right),\) \(\left( {2,6} \right)\) and \(\left( {7,1} \right).\)

    Example 19

    Find the area inside the ellipse \({\large\frac{{{x^2}}}{{{a^2}}}\normalsize} + {\large\frac{{{y^2}}}{{{b^2}}}\normalsize} = 1.\)

    Example 1.

    Calculate the derivative of the function \(g\left( x \right) = \int\limits_1^x {\sqrt {{t^3} + 4t} dt} \) at \(x = 2.\)

    Solution.

    We apply the Fundamental Theorem of Calculus, Part \(1:\)

    \[{g^\prime\left( x \right) }={ \frac{d}{{dx}}\left( {\int\limits_a^x {f\left( t \right)dt} } \right) }={ f\left( x \right).}\]

    Hence

    \[{g^\prime\left( x \right) }={ \frac{d}{{dx}}\left( {\int\limits_1^x {\sqrt {{t^3} + 4t} dt} } \right) }={ \sqrt {{x^3} + 4x} .}\]

    Substituting \(x = 2\) yields

    \[{g^\prime\left( 2 \right) }={ \sqrt {{2^3} + 4 \cdot 2} }={ \sqrt {16} }={ 4.}\]

    Example 2.

    Calculate the derivative of the function \(g\left( x \right) = \int\limits_{ – \large{\frac{\pi }{2}}\normalsize}^x {\sqrt {{{\sin }^2}t + 2} dt} \) at \(x = \large{\large{\frac{\pi }{6}}\normalsize}.\)

    Solution.

    We use the Fundamental Theorem of Calculus, Part \(1:\)

    \[{g^\prime\left( x \right) }={ \frac{d}{{dx}}\left( {\int\limits_a^x {f\left( t \right)dt} } \right) }={ f\left( x \right).}\]

    Then

    \[{g^\prime\left( x \right) }={ \frac{d}{{dx}}\left( {\int\limits_{ – \frac{\pi }{2}}^x {\sqrt {{{\sin }^2}t + 2} dt} } \right) }={ \sqrt {{{\sin }^2}x + 2} .}\]

    Note that the lower limit of integration \({ – \large{\frac{\pi }{2}}\normalsize}\) does not affect the answer.

    Now we compute the value of the derivative for \(x = \large{\frac{\pi }{6}}\normalsize :\)

    \[{g^\prime\left( {\frac{\pi }{6}} \right) }={ \sqrt {{{\sin }^2}\frac{\pi }{6} + 2} }={ \sqrt {{{\left( {\frac{1}{2}} \right)}^2} + 2} }={ \sqrt {\frac{9}{4}} }={ \frac{3}{2}.}\]

    Example 3.

    Find the derivative of the function \(g\left( x \right) = \int\limits_3^{{x^2}} {\large{\frac{{dt}}{t}}\normalsize}.\)

    Solution.

    We introduce the new function

    \[{h\left( u \right) = \int\limits_3^u {\frac{{dt}}{t}}.}\]

    Using the FTC1, we have

    \[{h^\prime\left( u \right) }={ \frac{1}{u}.}\]

    As \(g\left( x \right) = h\left( {{x^2}} \right),\) then by the chain rule

    \[{g^\prime\left( x \right) = \left[ {h\left( {{x^2}} \right)} \right]^\prime }={ h^\prime\left( {{x^2}} \right) \cdot \left( {{x^2}} \right)^\prime }={ h^\prime\left( {{x^2}} \right) \cdot 2x }={ \frac{1}{{{x^2}}} \cdot 2x }={ \frac{2}{x}}\]

    Example 4.

    Find the derivative of the function \(g\left( x \right) = \int\limits_0^{{x^2}} {\sqrt {1 + {t^2}} dt}.\)

    Solution.

    Since the upper limit of integration is not \(x,\) we apply the chain rule. Let \(u = {x^2},\) then \(u^\prime = 2x.\)

    Consider the new function

    \[h\left( u \right) = \int\limits_0^u {\sqrt {1 + {t^2}} dt} .\]

    By the FTC1, we can write

    \[h^\prime\left( u \right) = \sqrt {1 + {u^2}} .\]

    As \(g\left( x \right) = h\left( {{x^2}} \right),\) we have

    \[{g^\prime\left( x \right) = \left[ {h\left( {{x^2}} \right)} \right]^\prime }={ h^\prime\left( {{x^2}} \right) \cdot \left( {{x^2}} \right)^\prime }={ \sqrt {1 + {{\left( {{x^2}} \right)}^2}} \cdot 2x }={ 2x\sqrt {1 + {x^4}} .}\]

    Example 5.

    Find the derivative of the function \(g\left( x \right) = \int\limits_1^{{x^3}} {{t^2}dt}.\)

    Solution.

    Let \(u = {x^3},\) then \(u^\prime = 3{x^2}.\)

    We introduce the new function

    \[h\left( u \right) = \int\limits_0^u {{t^2}dt} .\]

    Using the FTC1, we obtain

    \[h^\prime\left( u \right) = {u^2}.\]

    Since \(g\left( x \right) = h\left( {{x^3}} \right),\) we have

    \[{g^\prime\left( x \right) = \left[ {h\left( {{x^3}} \right)} \right]^\prime }={ h^\prime\left( {{x^3}} \right) \cdot \left( {{x^3}} \right)^\prime }={ {\left( {{x^3}} \right)^2} \cdot 3{x^2} }={ {x^6} \cdot 3{x^2} }={ 3{x^8}.}\]

    Example 6.

    Find the derivative of the function \(g\left( x \right) = \int\limits_{{x^2}}^{{x^3}} {tdt}.\)

    Solution.

    We split the interval of integration \(\left[ {{x^2},{x^3}} \right]\) using an intermediate point \(c,\) so that \(c \in \left[ {{x^2},{x^3}} \right].\) Hence the derivative of \(g\left( x \right)\) is written in the form

    \[{g^\prime\left( x \right) }={ \frac{d}{{dx}}\left( {\int\limits_{{x^2}}^{{x^3}} {tdt} } \right) }={ \frac{d}{{dx}}\left( {\int\limits_{{x^2}}^c {tdt} + \int\limits_c^{{x^3}} {tdt} } \right) }={ \frac{d}{{dx}}\left( {\int\limits_c^{{x^3}} {tdt} – \int\limits_c^{{x^2}} {tdt} } \right) }={ \frac{d}{{dx}}\left( {\int\limits_c^{{x^3}} {tdt} } \right) – \frac{d}{{dx}}\left( {\int\limits_c^{{x^2}} {tdt} } \right).}\]

    We calculate both terms using the FTC1 and the chain rule:

    \[{\frac{d}{{dx}}\left( {\int\limits_c^{{x^3}} {tdt} } \right) }={ {x^3} \cdot \left( {{x^3}} \right)^\prime }={ {x^3} \cdot 3{x^2} }={ 3{x^5};}\]

    \[{\frac{d}{{dx}}\left( {\int\limits_c^{{x^2}} {tdt} } \right) }={ {x^2} \cdot \left( {{x^2}} \right)^\prime }={ {x^2} \cdot 2x }={ 2{x^3}.}\]

    Then

    \[g^\prime\left( x \right) = 3{x^5} – 2{x^3}.\]

    Example 7.

    Calculate the derivative of the function \(g\left( x \right) = \int\limits_{\sqrt x }^x {\left( {{t^2} – t} \right)dt} \) at \(x = 1.\)

    Solution.

    We split the integral function into two terms:

    \[{g\left( x \right) }={ \int\limits_{\sqrt x }^x {\left( {{t^2} – t} \right)dt} }={ \int\limits_{\sqrt x }^c {\left( {{t^2} – t} \right)dt} + \int\limits_c^x {\left( {{t^2} – t} \right)dt} }={ \int\limits_c^x {\left( {{t^2} – t} \right)dt} – \int\limits_c^{\sqrt x } {\left( {{t^2} – t} \right)dt},}\]

    where \(c \in \left[ {{x^2},{x^3}} \right].\)

    Find the derivative of \(g\left( x \right)\) using the FTC1 and the chain rule (for the second term):

    \[{\frac{d}{{dx}}\int\limits_c^x {\left( {{t^2} – t} \right)dt} }={ {x^2} – x;}\]

    \[{\frac{d}{{dx}}\int\limits_c^{\sqrt x } {\left( {{t^2} – t} \right)dt} }={ \left( {{{\left( {\sqrt x } \right)}^2} – \sqrt x } \right) \cdot \left( {\sqrt x } \right)^\prime }={ \left( {x – \sqrt x } \right) \cdot \frac{1}{{2\sqrt x }} }={ \frac{{\sqrt x }}{2} – \frac{1}{2}.}\]

    Then

    \[{g^\prime\left( x \right) }={ \left( {{x^2} – x} \right) }-{ \left( {\frac{{\sqrt x }}{2} – \frac{1}{2}} \right) }={ {x^2} – x – \frac{{\sqrt x }}{2} + \frac{1}{2}.}\]

    At the point \(x = 1,\) the derivative is equal to

    \[{g^\prime\left( 1 \right) }={ {1^2} – 1 }-{ \frac{{\sqrt 1 }}{2} + \frac{1}{2} }={ 0.}\]

    Example 8.

    Evaluate the integral \(\int\limits_0^2 {\left( {{x^3} – {x^2}} \right)dx}.\)

    Solution.

    Using the Fundamental Theorem of Calculus, Part \(2,\) we have

    \[
    {\int\limits_0^2 {\left( {{x^3} – {x^2}} \right)dx} }
    = {\left. {\left( {\frac{{{x^4}}}{4} – \frac{{{x^3}}}{3}} \right)} \right|_0^2 }
    = {\left( {\frac{{16}}{4} – \frac{8}{3}} \right) – 0 }={ \frac{4}{3}.}
    \]

    Example 9.

    Evaluate the integral \(\int\limits_{ – 1}^1 {\left( {{t^2} + {t^{21}}} \right)dt}.\)

    Solution.

    An antiderivative of the function \({{t^2} + {t^{21}}}\) is \(\large{\frac{{{t^3}}}{3} + \frac{{{t^{22}}}}{{22}}}\normalsize.\) Then using the Fundamental Theorem of Calculus, Part \(2,\) we have

    \[\require{cancel}{\int\limits_{ – 1}^1 {\left( {{t^2} + {t^{21}}} \right)dt} }={ \left. {\left[ {\frac{{{t^3}}}{3} + \frac{{{t^{22}}}}{{22}}} \right]} \right|_{ – 1}^1 }={ \left( {\frac{{{1^3}}}{3} + \frac{{{1^{22}}}}{{22}}} \right) }-{ \left( {\frac{{{{\left( { – 1} \right)}^3}}}{3} + \frac{{{{\left( { – 1} \right)}^{22}}}}{{22}}} \right) }={ \frac{1}{3} + \cancel{\frac{1}{{22}}} + \frac{1}{3} – \cancel{\frac{1}{{22}}} }={ \frac{2}{3}.}\]

    Example 10.

    Calculate the integral \(\int\limits_0^1 {\left( {\sqrt[\large 3\normalsize]{t} – \sqrt t } \right)dt}.\)

    Solution.

    \[
    {\int\limits_0^1 {\left( {\sqrt[\large 3\normalsize]{t} – \sqrt t } \right)dt} }
    = {\int\limits_0^1 {\left( {{t^{\large\frac{1}{3}\normalsize}} – {t^{\large\frac{1}{2}\normalsize}}} \right)dt} }
    = {\left. {\left( {\frac{{{t^{\large\frac{1}{3}\normalsize + 1}}}}{{\frac{1}{3} + 1}} – \frac{{{t^{\large\frac{1}{2}\normalsize + 1}}}}{{\frac{1}{2} + 1}}} \right)} \right|_0^1 }
    = {\left. {\left( {\frac{{3{t^{\large\frac{4}{3}\normalsize}}}}{4} – \frac{{2{t^{\large\frac{3}{2}\normalsize}}}}{3}} \right)} \right|_0^1 }
    = {\left( {\frac{3}{4} – \frac{2}{3}} \right) – 0 }={ \frac{1}{{12}}.}
    \]

    Example 11.

    Evaluate the integral \(\int\limits_0^1 {{\large\frac{x}{{{{\left( {3{x^2} – 1} \right)}^4}}}\normalsize} dx}.\)

    Solution.

    First we make the substitution:

    \[
    {t = 3{x^2} – 1,\;\;}\Rightarrow
    {dt = 6xdx,\;\;}\Rightarrow
    {xdx = \frac{{dt}}{6}.}
    \]

    Determine the new limits of integration. When \(x = 0,\) then \(t = -1.\) When \(x = 1,\) then we have \(t = 2.\) So, the integral with the new variable \(t\) can be easily calculated:

    \[
    {\int\limits_0^1 {\frac{x}{{{{\left( {3{x^2} – 1} \right)}^4}}}dx} }
    = {\int\limits_{ – 1}^2 {\frac{{\frac{{dt}}{6}}}{{{t^4}}}} }
    = {\frac{1}{6}\int {{t^{ – 4}}dt} }
    = {\frac{1}{6}\left. {\left( {\frac{{{t^{ – 3}}}}{{ – 3}}} \right)} \right|_{ – 1}^2 }
    = { – \frac{1}{{18}}\left( {\frac{1}{8} – 1} \right) }
    = {\frac{7}{{144}}.}
    \]

    Example 12.

    Evaluate the integral \(\int\limits_1^e {\left( {t + \large{\frac{1}{t}}\normalsize} \right)dt}.\)

    Solution.

    An antiderivative of the function \({t + \large{\frac{1}{t}}\normalsize}\) has the form \(\large{\frac{{{t^2}}}{2}}\normalsize + \ln t.\) Hence, by the Fundamental Theorem, Part \(2,\) we have

    \[{\int\limits_1^e {\left( {t + \frac{1}{t}} \right)dt} }={ \left. {\left[ {\frac{{{t^2}}}{2} + \ln t} \right]} \right|_1^e }={ \left( {\frac{{{e^2}}}{2} + \ln e} \right) }-{ \left( {\frac{{{1^2}}}{2} + \ln 1} \right) }={ \frac{{{e^2}}}{2} + 1 – \frac{1}{2} – 0 }={ \frac{{{e^2}}}{2} + \frac{1}{2}.}\]

    Example 13.

    Evaluate the integral \(\int\limits_0^{\ln 2} {x{e^{ – x}}dx}.\)

    Solution.

    We can write

    \[
    {I = \int\limits_0^{\ln 2} {x{e^{ – x}}dx} }
    = { – \int\limits_0^{\ln 2} {xd\left( {{e^{ – x}}} \right)} .}
    \]

    Apply integration by parts: \({\large\int\normalsize} {udv} \) \(= uv – {\large\int\normalsize} {vdu} .\) In this case, let

    \[
    {u = x,\;\;}\kern-0.3pt{dv = d\left( {{e^{ – x}}} \right),\;\;}\Rightarrow
    {du = 1,\;\;}\kern-0.3pt{v = {e^{ – x}}.}
    \]

    Hence, the integral is

    \[
    {I = – \int\limits_0^{\ln 2} {xd\left( {{e^{ – x}}} \right)} }
    = { – \left[ {\left. {\left( {x{e^{ – x}}} \right)} \right|_0^{\ln 2} – \int\limits_0^{\ln 2} {{e^{ – x}}dx} } \right] }
    = {{ – \left. {\left( {x{e^{ – x}}} \right)} \right|_0^{\ln 2} }+{ \int\limits_0^{\ln 2} {{e^{ – x}}dx} }}
    = {{ – \left. {\left( {x{e^{ – x}}} \right)} \right|_0^{\ln 2} }-{ \left. {\left( {{e^{ – x}}} \right)} \right|_0^{\ln 2} }}
    = { – \left. {\left[ {{e^{ – x}}\left( {x + 1} \right)} \right]} \right|_0^{\ln 2} }
    = {{ – {e^{ – \ln 2}}\left( {\ln 2 + 1} \right) }+{ {e^0} \cdot 1 }}
    = { – \frac{{\ln 2}}{2} – \frac{{\ln e}}{2} + \ln e }
    = {\frac{{\ln e}}{2} – \frac{{\ln 2}}{2} }
    = {\frac{1}{2}\left( {\ln e – \ln 2} \right) }
    = {\frac{1}{2}\ln \frac{e}{2}.}
    \]

    Example 14.

    Evaluate the integral \(\int\limits_{ – 1}^1 {\left| {x – \large{\frac{1}{2}}\normalsize} \right|dx}.\)

    Solution.

    The area enclosed by y=|x-1/2| is split into two pieces
    Figure 4.

    We rewrite the absolute value expression in the form

    \[\left| {x – \frac{1}{2}} \right| = \begin{cases} – x + \frac{1}{2}, & \text{if }x \lt \frac{1}{2}\\ x – \frac{1}{2}, & \text{if }x \ge \frac{1}{2} \end{cases}\]

    and split the interval of integration into two intervals such that

    \[{\int\limits_{ – 1}^1 {\left| {x – \frac{1}{2}} \right|dx} }={ \int\limits_{ – 1}^{\frac{1}{2}} {\left| {x – \frac{1}{2}} \right|dx} }+{ \int\limits_{\frac{1}{2}}^1 {\left| {x – \frac{1}{2}} \right|dx} }={ \int\limits_{ – 1}^{\frac{1}{2}} {\left( { – x + \frac{1}{2}} \right)dx} }+{ \int\limits_{\frac{1}{2}}^1 {\left( {x – \frac{1}{2}} \right)dx} .}\]

    Now we can apply the Fundamental Theorem of Calculus, Part \(2,\) to each of the integrals:

    \[{\int\limits_{ – 1}^1 {\left| {x – \frac{1}{2}} \right|dx} }={ \int\limits_{ – 1}^{\frac{1}{2}} {\left( { – x + \frac{1}{2}} \right)dx} }+{ \int\limits_{\frac{1}{2}}^1 {\left( {x – \frac{1}{2}} \right)dx} }={ \left[ { – \frac{{{x^2}}}{2} + \frac{x}{2}} \right]_{ – 1}^{\frac{1}{2}} }+{ \left[ {\frac{{{x^2}}}{2} – \frac{x}{2}} \right]_{\frac{1}{2}}^1 }={ \left[ {\left( { – \frac{1}{8} + \frac{1}{4}} \right) – \left( { – \frac{1}{2} – \frac{1}{2}} \right)} \right] }+{ \left[ {\left( {\cancel{\frac{1}{2}} – \cancel{\frac{1}{2}}} \right) – \left( {\frac{1}{8} – \frac{1}{4}} \right)} \right] }={ \frac{9}{8} + \frac{1}{8} }={ \frac{{10}}{8} }={ \frac{5}{4}.}\]

    Example 15.

    Evaluate the integral \(\int\limits_{ – 2}^1 {\left| {{x^2} – 1} \right|dx}.\)

    Solution.

    The area enclosed by y=|x^2-1| is split into two pieces
    Figure 5.

    We represent the absolute value expression as follows:

    \[{\left| {{x^2} – 1} \right| \text{ = }}\kern0pt{\begin{cases} {{x^2} – 1}, & \text{if }x \in \left( { – \infty , – 1} \right] \cup \left[ {1,\infty } \right)\\ 1 – {x^2}, & \text{if }x \in \left( { – 1,1} \right) \end{cases}}\]

    So we can split the initial integral into two integrals:

    \[{\int\limits_{ – 2}^1 {\left| {{x^2} – 1} \right|dx} }={ \int\limits_{ – 2}^{ – 1} {\left| {{x^2} – 1} \right|dx} }+{ \int\limits_{ – 1}^1 {\left| {{x^2} – 1} \right|dx} }={ \int\limits_{ – 2}^{ – 1} {\left( {{x^2} – 1} \right)dx} }+{ \int\limits_{ – 1}^1 {\left( {1 – {x^2}} \right)dx} .}\]

    Using the Fundamental Theorem of Calculus, Part \(2,\) we obtain:

    \[{\int\limits_{ – 2}^1 {\left| {{x^2} – 1} \right|dx} }={ \int\limits_{ – 2}^{ – 1} {\left( {{x^2} – 1} \right)dx} }+{ \int\limits_{ – 1}^1 {\left( {1 – {x^2}} \right)dx} }={ \left[ {\frac{{{x^3}}}{3} – x} \right]_{ – 2}^{ – 1} }+{ \left[ {x – \frac{{{x^3}}}{3}} \right]_{ – 1}^1 }={ \left[ {\left( { – \frac{1}{3} – \left( { – 1} \right)} \right) }\right.}-{\left.{ \left( { – \frac{8}{3} – \left( { – 2} \right)} \right)} \right] }+{ \left[ {\left( {1 – \frac{1}{3}} \right) – \left( { – 1 – \left( { – \frac{1}{3}} \right)} \right)} \right] }={ \frac{2}{3} + \frac{2}{3} + \frac{2}{3} + \frac{2}{3} }={ \frac{8}{3}.}\]

    Example 16.

    Find the area bounded by the curves \(y = {x^2}\) and \(y = \sqrt x.\)

    Solution.

    First we find the points of intersection (see Figure \(6\)):

    \[
    {{x^2} = \sqrt x ,\;\;}\Rightarrow
    {{x^2} – \sqrt x = 0,\;\;}\Rightarrow
    {\sqrt x \left( {{x^{\large\frac{3}{2}\normalsize}} – 1} \right) = 0,\;\;}\Rightarrow
    {{x_1} = 0,\;{x_2} = 1.}
    \]

    Area bounded by the curves y=x^2 and y=sqrt(x)
    Figure 6.

    As you can see, the curves intercept at the points \(\left( {0,0} \right)\) and \(\left( {1,1}\right).\) Hence, the area is given by

    \[
    {S = \int\limits_0^1 {\left( {\sqrt x – {x^2}} \right)dx} }
    = {\left. {\left( {\frac{{{x^{\large\frac{1}{2}\normalsize + 1}}}}{{\frac{1}{2} + 1}} – \frac{{{x^3}}}{3}} \right)} \right|_0^1 }
    = {\frac{1}{3}\left. {\left( {2\sqrt {{x^3}} – {x^3}} \right)} \right|_0^1 }={ \frac{1}{3}.}
    \]

    Example 17.

    Find the area bounded by the curves \(y = 2x – {x^2}\) and \(x + y = 0.\)

    Solution.

    First we find the points of intersection of the curves (see Figure \(7\)):

    \[
    {2x – {x^2} = – x,\;\;}\Rightarrow
    {{x^2} – 3x = 0,\;\;}\Rightarrow
    {x\left( {x – 3} \right) = 0,\;\;}\Rightarrow
    {{x_1} = 0,\;{x_2} = 3.}
    \]

    Area bounded by a parabola and a straight line
    Figure 7.

    The upper boundary of the region is the parabola \(y = 2x – {x^2},\) and the lower boundary is the straight line \(y = -x.\)

    The area is given by

    \[
    {S = \int\limits_0^3 {\left[ {2x – {x^2} – \left( { – x} \right)} \right]dx} }
    = {\int\limits_0^3 {\left( {2x – {x^2} + x} \right)dx} }
    = {\left. {\left( {{x^2} – \frac{{{x^3}}}{3} + \frac{{{x^2}}}{2}} \right)} \right|_0^3 }
    = {\left. {\left( {\frac{{3{x^2}}}{2} – \frac{{{x^3}}}{3}} \right)} \right|_0^3 }
    = {\frac{{27}}{2} – \frac{{27}}{3} }={ \frac{9}{2}.}
    \]

    Example 18.

    Find the area of the triangle with vertices at \(\left( {0,0} \right),\) \(\left( {2,6} \right)\) and \(\left( {7,1} \right).\)

    Solution.

    First we find an equation of the side \(OA\) (Figure \(8\)):

    \[
    {\frac{{x – {x_O}}}{{{x_A} – {x_O}}} = \frac{{y – {y_O}}}{{{y_A} – {y_O}}},\;\;}\Rightarrow
    {\frac{{x – 0}}{{2 – 0}} = \frac{{y – 0}}{{6 – 0}},\;\;}\Rightarrow
    {\frac{x}{2} = \frac{y}{6},\;\;}\Rightarrow
    {y = 3x.}
    \]

    Area of a triangle
    Figure 8.

    Similarly, we find an equation of the side \(OB:\)

    \[
    {\frac{{x – {x_O}}}{{{x_B} – {x_O}}} = \frac{{y – {y_O}}}{{{y_B} – {y_O}}},\;\;}\Rightarrow
    {\frac{{x – 0}}{{7 – 0}} = \frac{{y – 0}}{{1 – 0}},\;\;}\Rightarrow
    {\frac{x}{7} = \frac{y}{1},\;\;}\Rightarrow
    {y = \frac{x}{7}.}
    \]

    Next, find an equation of the side \(AB:\)

    \[
    {\frac{{x – {x_B}}}{{{x_A} – {x_B}}} = \frac{{y – {y_B}}}{{{y_A} – {y_B}}},\;\;}\Rightarrow
    {\frac{{x – 2}}{{7 – 2}} = \frac{{y – 6}}{{1 – 6}},\;\;}\Rightarrow
    {\frac{{x – 2}}{5} = \frac{{y – 6}}{{ – 5}},\;\;}\Rightarrow
    {y = 8 – x.}
    \]

    As you can see from Figure \(8,\) the area of the this triangle can be calculated as the sum of two integrals:

    \[
    {S = {I_1} + {I_2} }
    = {\int\limits_0^2 {\left( {3x – \frac{x}{7}} \right)dx} }+{ \int\limits_2^7 {\left( {8 – x – \frac{x}{7}} \right)dx} }
    = {\left. {\left( {\frac{{10{x^2}}}{7}} \right)} \right|_0^2 }+{ \left. {\left( {8x – \frac{{4{x^2}}}{7}} \right)} \right|_2^7 }
    = {\frac{{10 \cdot 4}}{7} + \left( {56 – \frac{{4 \cdot 49}}{7}} \right) }-{ \left( {16 – \frac{{4 \cdot 4}}{7}} \right) }={ 20.}
    \]

    Example 19.

    Find the area inside the ellipse \({\large\frac{{{x^2}}}{{{a^2}}}\normalsize} + {\large\frac{{{y^2}}}{{{b^2}}}\normalsize} = 1.\)

    Solution.

    By symmetry (see Figure \(9\)), the area of the ellipse is twice the area above the \(x\)-axis.

    Area of the ellipse x^2/a^2+y^2/b^2=1
    Figure 9.

    The latter is given by

    \[
    {{S_{\frac{1}{2}}} }={ \int\limits_{ – a}^a {\sqrt {{b^2}\left( {1 – \frac{{{x^2}}}{{{a^2}}}} \right)} dx} }
    = {\frac{b}{a}\int\limits_{ – a}^a {\sqrt {{a^2} – {x^2}} dx} .}
    \]

    To calculate the last integral, we use the trigonometric substitution \(x = a\sin t,\) \(dx = a\cos tdt.\)

    Refine the limits of integration. When \(x = -a,\) then \(\sin t = -1\) and \(t = – {\large\frac{\pi }{2}\normalsize}.\) When \(x = a,\) then \(\sin t = 1\) and \(t = {\large\frac{\pi }{2}\normalsize}.\) Thus we get

    \[
    {{S_{\frac{1}{2}}} }={ \frac{b}{a}\int\limits_{ – a}^a {\sqrt {{a^2} – {x^2}} dx} }
    = {\frac{b}{a}\int\limits_{ – \large\frac{\pi }{2}\normalsize}^{\large\frac{\pi }{2}\normalsize} {\sqrt {{a^2} – {a^2}{{\sin }^2}t}\, }}\kern0pt{{ a\cos tdt} }
    = {ab\int\limits_{ – \large\frac{\pi }{2}\normalsize}^{\large\frac{\pi }{2}\normalsize} {{{\cos }^2}tdt} }
    = {ab\int\limits_{ – \large\frac{\pi }{2}\normalsize}^{\large\frac{\pi }{2}\normalsize} {\frac{{1 + \cos 2t}}{2}dt} }
    = {\frac{{ab}}{2}\int\limits_{ – \large\frac{\pi }{2}\normalsize}^{\large\frac{\pi }{2}\normalsize} {\left( {1 + \cos 2t} \right)dt} }
    = {\frac{{ab}}{2}\left. {\left( {t + \frac{{\sin 2t}}{2}} \right)} \right|_{ – \large\frac{\pi }{2}\normalsize}^{\large\frac{\pi }{2}\normalsize} }
    = {\frac{{ab}}{2}\left[ {\frac{\pi }{2} + \frac{{\sin \pi }}{2} }\right.}-{\left.{ \left( { – \frac{\pi }{2}} \right) – \frac{{\sin \left( { – \pi } \right)}}{2}} \right] }
    = {\frac{{\pi ab}}{2}.}
    \]

    Hence, the total area of the ellipse is \(\pi ab.\)