# The Fundamental Theorem of Calculus

## Solved Problems

Click or tap a problem to see the solution.

### Example 9

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

### Example 10

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

### Example 11

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

### Example 12

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

### Example 13

Evaluate the integral $\int\limits_0^{\ln 2} {x{e^{ - x}}dx}.$

### Example 14

Evaluate the integral $\int\limits_{ - 1}^1 {\left| {x - \frac{1}{2}} \right|dx}.$

### Example 15

Evaluate the integral $\int\limits_{ - 2}^1 {\left| {{x^2} - 1} \right|dx}.$

### Example 16

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

### Example 17

Find the area bounded by the curves $y = 2x - {x^2} \text{ and } x + y = 0.$

### 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 $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1.$

### 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 $$\frac{{{t^3}}}{3} + \frac{{{t^{22}}}}{{22}}.$$ Then using the Fundamental Theorem of Calculus, Part $$2,$$ we have

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

### Example 10.

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

Solution.

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

### Example 11.

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

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 $\int\limits_1^e {\left( {t + \frac{1}{t}} \right)dt}.$

Solution.

An antiderivative of the function $${t + \frac{1}{t}}$$ has the form $$\frac{{{t^2}}}{2} + \ln t.$$ 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: $$\int {udv}$$ $$= uv - \int {vdu} .$$ In this case, let

$u = x,\;\; dv = d\left( {{e^{ - x}}} \right),\;\; \Rightarrow du = 1,\;\; v = {e^{ - x}}.$

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 $\int\limits_{ - 1}^1 {\left| {x - \frac{1}{2}} \right|dx}.$

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:

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

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:

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

### Example 16.

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

Solution.

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

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

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

$S = \int\limits_0^1 {\left( {\sqrt x - {x^2}} \right)dx} = \left. {\left( {\frac{{{x^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}} - \frac{{{x^3}}}{3}} \right)} \right|_0^1 = \frac{1}{3}\left. {\left( {2\sqrt {{x^3}} - {x^3}} \right)} \right|_0^1 = \frac{1}{3}.$

### Example 17.

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

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.

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 $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1.$

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 $$t = - {\frac{\pi }{2}}.$$ When $$x = a,$$ then $$\sin t = 1$$ and $$t = {\frac{\pi }{2}}.$$ Thus we get

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

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