Activity 56.
Use the Fundamental Theorem of Calculus to evaluate each of the following integrals exactly. For each, sketch a graph of the integrand on the relevant interval and write one sentence that explains the meaning of the value of the integral in terms of the (net signed) area bounded by the curve.
(a)
\(\int_{-1}^4 (2-2x) \, dx\)
Hint.
Find a function whose derivative is \(2 - 2x\text{.}\)
Answer.
\(\int_{-1}^4 (2-2x) \, dx = -5\text{.}\)
Solution.
Because \(\frac{d}{dx}[2x - x^2] = 2-2x\text{,}\) by the Fundamental Theorem of Calculus,
\begin{equation*}
\int_{-1}^4 (2-2x) \, dx = \left. 2x - x^2 \right|_{-1}^4\text{,}
\end{equation*}
and therefore
\begin{equation*}
\int_{-1}^4 (2-2x) \, dx = (2 \cdot 4 - 4^2) - (2(-1) - (-1)^2) = -8 + 3 = -5\text{.}
\end{equation*}
(b)
\(\int_{0}^{\frac{\pi}{2}} \sin(x) \, dx\)
Hint.
Which familiar function has derivative \(\sin(x)\text{?}\)
Answer.
\(\int_{0}^{\frac{\pi}{2}} \sin(x) \, dx = 1\text{.}\)
Solution.
Since \(\frac{d}{dx} [\cos(x)] = -\sin(x)\text{,}\) an antiderivative of \(f(x) = \sin(x)\) is \(F(x) = -\cos(x)\text{.}\) Therefore, by the FTC,
\begin{equation*}
\int_{0}^{\frac{\pi}{2}} \sin(x) \, dx = \left. -\cos(x) \right|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\text{,}
\end{equation*}
so
\begin{equation*}
\int_{0}^{\frac{\pi}{2}} \sin(x) \, dx = -\cos(\frac{\pi}{2}) - (-\cos(0)) = -0 + 1 = 1\text{.}
\end{equation*}
(c)
\(\int_0^1 e^x \, dx\)
Hint.
What is special about \(e^x\) when it comes to differentiation?
Answer.
\(\int_0^1 e^x \, dx = e-1\text{.}\)
Solution.
Since \(e^x\) is its own derivative, it is also its own antiderivative. Hence,
\begin{equation*}
\int_0^1 e^x \, dx = \left. e^x \right|_0^1 = e^1 - e^0 = e-1\text{.}
\end{equation*}
(d)
\(\int_{-1}^{1} x^5 \, dx\)
Hint.
Consider the derivative of \(x^6\text{.}\)
Answer.
\(\int_{-1}^{1} x^5 \, dx = 0\text{.}\)
Solution.
Note that since \(\frac{d}{dx} [x^6] = 6x^5\text{,}\) it follows that \(\frac{d}{dx} [\frac{1}{6}x^6] = x^5\text{,}\) and thus
\begin{equation*}
\int_{-1}^{1} x^5 \, dx = \left. \frac{1}{6}x^6 \right|_{-1}^1 = \frac{1}{6} (1)^6 - \frac{1}{6}(-1)^6 = 0\text{.}
\end{equation*}
(e)
\(\int_0^2 (3x^3 - 2x^2 - e^x) \, dx\)
Hint.
Find an antiderivative for each of the three individual terms in the integrand.
Answer.
\(\int_0^2 (3x^3 - 2x^2 - e^x) \, dx = \frac{23}{3} - e^2\text{.}\)
Solution.
Using the sum and constant multiple rules for differentiation, we can see that similar results hold for antidifferentiation, and thus that \(F(x) = \frac{3}{4} x^4 - \frac{2}{3} x^3 - e^x\) is an antiderivative of \(f(x) = 3x^3 - 2x^2 - e^x\text{.}\) Now, by the FTC,
\begin{align*}
\int_0^2 (3x^3 - 2x^2 - e^x) \, dx =\mathstrut \amp \left. \frac{3}{4} x^4 - \frac{2}{3} x^3 - e^x \right|_0^2\\
=\mathstrut \amp \frac{3}{4} (2)^4 - \frac{2}{3} (2)^3 - e^2 - (\frac{3}{4} (0)^4 - \frac{2}{3} (0)^3 - e^0)\\
=\mathstrut \amp 12 - \frac{16}{3} - e^2 - (0 - 0 - 1)\\
=\mathstrut \amp \frac{23}{3} - e^2\text{.}
\end{align*}
