Calculus

Integration of Functions

Integration of Functions Logo

The Definite Integral and Fundamental Theorem of Calculus

  • Let \(f\left( x \right)\) be a continuous function on the closed interval \(\left[ {a,b} \right].\) Divide this interval into \(n\) partial subintervals. Choose a sample point \({\xi_i}\) in each subinterval and make up an integral sum \(\sum\limits_{i = 1}^n {f\left( {{\xi _i}} \right)\Delta {x_i}} \), where \(\Delta {x_i}\) is the length of the \(i\)th subinterval.

    The definite integral of \(f\left( x \right)\) from \(a\) to \(b\) is defined to be the limit

    \[\require{AMSmath.js} {\int\limits_a^b {f\left( x \right)dx} }={ \lim\limits_{\substack{ n \to \infty \\ \text{max}\,\Delta {x_i} \to 0}} \sum\limits_{i = 1}^n {f\left( {{\xi _i}} \right)\Delta {x_i}} ,}\]

    where \(\Delta {x_i} = {x_i} – {x_{i – 1}},\) \({x_{i – 1}} \le {\xi _i} \le {x_i}.\)

    Properties of the Definite Integral

    We assume below that \(f\left( x \right)\) and \(g\left( x \right)\) are continuous functions on the closed interval \(\left[ {a,b} \right].\)

    1. \(\int\limits_a^b {1dx} = b – a\)
    2. \(\int\limits_a^b {kf\left( x \right)dx} \) \(= k \int\limits_a^b {f\left( x \right)dx} ,\) where \(k\) is a constant;
    3. \(\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]dx} \) \(= \int\limits_a^b {f\left( x \right)dx} \) \(+{ \int\limits_a^b {g\left( x \right)dx} }\)
    4. \(\int\limits_a^b {f\left( x \right)dx} \) \(= \int\limits_a^c {f\left( x \right)dx} \) \(+ \int\limits_c^b {f\left( x \right)dx} ,\) where \(a \lt c \lt b;\)
    5. If \(0 \le f\left( x \right) \le g\left( x \right)\) for all \(x \in \left[ {a,b} \right],\) then \(0 \le \int\limits_a^b {f\left( x \right)dx} \) \(\le \int\limits_a^b {g\left( x \right)dx} .\)
    6. \(\int\limits_a^a {f\left( x \right)dx} = 0\)
    7. \(\int\limits_a^b {f\left( x \right)dx} \) \(= – \int\limits_b^a {f\left( x \right)dx}\)
    8. If \(f\left( x \right) \ge 0\) in the interval \(\left[ {a,b} \right],\) then \(\int\limits_a^b {f\left( x \right)dx} \ge 0\)

    The Fundamental Theorem of Calculus

    Let \(f\left( x \right)\) be a function, which is continuous on the closed interval \(\left[ {a,b} \right].\) If \(F\left( x \right)\) is any antiderivative of \(f\left( x \right)\) on \(\left[ {a,b} \right],\) then

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

    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 \(x = a,\) \(x = b\) \(\left({\text{Figure }1}\right)\) 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 1.

    Let \(F\left( x \right)\) and \(G\left( x \right)\) be indefinite integrals 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 \(x = a,\) \(x = b\) \(\left({\text{Figure }2}\right)\) 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 2.

    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

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

    Example 2

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

    Example 3

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

    Example 4

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

    Example 5

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

    Example 6

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

    Example 7

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

    Example 8

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

    Example 1.

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

    Solution.

    Using the fundamental theorem of calculus, 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 2.

    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 3.

    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:

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

    Page 1
    Problems 1-3
    Page 2
    Problems 4-8