# 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 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

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

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

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

### 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 or tap 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.

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.

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 $$6$$):

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

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 $$7$$):

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

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 $$8$$):

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

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.

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 $$\pi ab.$$