The Fundamental Theorem of Calculus
Solved Problems
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.
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.
Figure 5. 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 \(6\)):
\[{x^2} = \sqrt x ,\;\; \Rightarrow {x^2} - \sqrt x = 0,\;\; \Rightarrow \sqrt x \left( {{x^{\frac{3}{2}}} - 1} \right) = 0,\;\; \Rightarrow {x_1} = 0,\;{x_2} = 1.\]
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
\[S = \int\limits_0^1 {\left( {\sqrt x - {x^2}} \right)dx} = \left. {\left( {\frac{{{x^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}} - \frac{{{x^3}}}{3}} \right)} \right|_0^1 = \frac{1}{3}\left. {\left( {2\sqrt {{x^3}} - {x^3}} \right)} \right|_0^1 = \frac{1}{3}.\]
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 \(7\)):
\[2x - {x^2} = - x,\;\; \Rightarrow {x^2} - 3x = 0,\;\; \Rightarrow x\left( {x - 3} \right) = 0,\;\; \Rightarrow {x_1} = 0,\;{x_2} = 3.\]
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.
We find an equation of the side \(OA\) (Figure \(8\)):
\[\frac{{x - {x_O}}}{{{x_A} - {x_O}}} = \frac{{y - {y_O}}}{{{y_A} - {y_O}}},\;\; \Rightarrow \frac{{x - 0}}{{2 - 0}} = \frac{{y - 0}}{{6 - 0}},\;\; \Rightarrow \frac{x}{2} = \frac{y}{6},\;\; \Rightarrow y = 3x.\]
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 \[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1.\]
Solution.
By symmetry (see Figure \(9\)), the area of the ellipse is twice the area above the \(x\)-axis.
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 \(t = - {\frac{\pi }{2}}.\) 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 \(\pi ab.\)
