Activity 57.
Use your knowledge of derivatives of basic functions to complete Table 195 of antiderivatives. For each entry, your task is to find a function \(F\) whose derivative is the given function \(f\text{.}\) When finished, use the FTC and the results in the table to evaluate the three given definite integrals.
| given function, \(f(x)\) | antiderivative, \(F(x)\) |
| \(k\text{,}\) (\(k\) is constant) | |
| \(x^n\text{,}\) \(n \ne -1\) | |
| \(\frac{1}{x}\text{,}\) \(x \gt 0\) | |
| \(\sin(x)\) | |
| \(\cos(x)\) | |
| \(\sec(x) \tan(x)\) | |
| \(\csc(x) \cot(x)\) | |
| \(\sec^2 (x)\) | |
| \(\csc^2 (x)\) | |
| \(e^x\) | |
| \(a^x\) \((a \gt 1)\) | |
| \(\frac{1}{1+x^2}\) | |
| \(\frac{1}{\sqrt{1-x^2}}\) |
(a)
\(\displaystyle \int_0^1 \left(x^3 - x - e^x + 2\right) \,dx\)
Hint.
Answer.
| given function, \(f(x)\) | antiderivative, \(F(x)\) |
| \(k\text{,}\) (\(k \ne 0\)) | \(kx\) |
| \(x^n\text{,}\) \(n \ne -1\) | \(\frac{1}{n+1}x^{n+1}\) |
| \(\frac{1}{x}\text{,}\) \(x \gt 0\) | \(\ln(x)\) |
| \(\sin(x)\) | \(-\cos(x)\) |
| \(\cos(x)\) | \(\sin(x)\) |
| \(\sec(x) \tan(x)\) | \(\sec(x)\) |
| \(\csc(x) \cot(x)\) | \(-\csc(x)\) |
| \(\sec^2 (x)\) | \(\tan(x)\) |
| \(\csc^2 (x)\) | \(-\cot(x)\) |
| \(e^x\) | \(e^x\) |
| \(a^x\) \((a \gt 1)\) | \(\frac{1}{\ln(a)} a^x\) |
| \(\frac{1}{1+x^2}\) | \(\arctan(x)\) |
| \(\frac{1}{\sqrt{1-x^2}}\) | \(\arcsin(x)\) |
\(\int_0^1 \left(x^3 - x - e^x + 2\right) \,dx = \frac{11}{4} - e\text{.}\)
Solution.
| given function, \(f(x)\) | antiderivative, \(F(x)\) |
| \(k\text{,}\) (\(k \ne 0\)) | \(kx\) |
| \(x^n\text{,}\) \(n \ne -1\) | \(\frac{1}{n+1}x^{n+1}\) |
| \(\frac{1}{x}\text{,}\) \(x \gt 0\) | \(\ln(x)\) |
| \(\sin(x)\) | \(-\cos(x)\) |
| \(\cos(x)\) | \(\sin(x)\) |
| \(\sec(x) \tan(x)\) | \(\sec(x)\) |
| \(\csc(x) \cot(x)\) | \(-\csc(x)\) |
| \(\sec^2 (x)\) | \(\tan(x)\) |
| \(\csc^2 (x)\) | \(-\cot(x)\) |
| \(e^x\) | \(e^x\) |
| \(a^x\) \((a \gt 1)\) | \(\frac{1}{\ln(a)} a^x\) |
| \(\frac{1}{1+x^2}\) | \(\arctan(x)\) |
| \(\frac{1}{\sqrt{1-x^2}}\) | \(\arcsin(x)\) |
By standard antiderivative rules and the FTC,
\begin{align*}
\int_0^1 \left(x^3 - x - e^x + 2\right) \,dx =\mathstrut \amp \left. \frac{1}{4} x^4 - \frac{1}{2}x^2 - e^x + 2x \right|_0^1\\
=\mathstrut \amp \left(\frac{1}{4} (1)^4 - \frac{1}{2}(1)^2 - e^1 + 2(1) \right) - \left(\frac{1}{4} (0)^4 - \frac{1}{2}(0)^2 - e^0 + 2(0) \right)\\
=\mathstrut \amp \frac{1}{4} - \frac{1}{2} - e + 2 - \left(0 - 0 - 1 + 0 \right)\\
=\mathstrut \amp -\frac{1}{4} - e + 3\\
=\mathstrut \amp \frac{11}{4} - e\text{.}
\end{align*}
(b)
\(\displaystyle \int_0^{\pi/3} (2\sin (t) - 4\cos(t) + \sec^2(t) - \pi) \, dt\)
Hint.
Answer.
\(\int_0^{\pi/3} (2\sin (t) - 4\cos(t) + \sec^2(t) - \pi) \, dt = 1 - \sqrt{3} - \frac{\pi^2}{3}\text{.}\)
Solution.
Calculating the needed antiderivative and applying the FTC,
\begin{align*}
\int_0^{\pi/3} (2\sin (t) - 4\cos(t) + \sec^2(t) - \pi) \, dt =\mathstrut \amp \left. \left(-2\cos(t) - 4\sin(t) + \tan(t) - \pi t \right) \right|_0^{\pi/3}\\
=\mathstrut \amp \left(-2\cos(\pi/3) - 4\sin(\pi/3) + \tan(\pi/3) - \pi (\pi/3) \right) -\\
\ \amp \left(-2\cos(0)) - 4\sin(0) + \tan(0) - \pi (0) \right)\\
=\mathstrut \amp -2 \cdot \frac{1}{2} - 4 \cdot \frac{\sqrt{3}}{2} + \sqrt{3} - \frac{\pi^2}{3} - (-2 - 0 + 0 - 0)\\
=\mathstrut \amp 1 - \sqrt{3} - \frac{\pi^2}{3}\text{.}
\end{align*}
(c)
\(\displaystyle \int_0^1 (\sqrt{x}-x^2) \, dx\)
Hint.
Answer.
\(\int_0^1 (\sqrt{x} - x^2) \, dx = \frac{1}{3}\text{.}\)
Solution.
Noting that \(\frac{d}{dx}[\frac{2}{3} x^{3/2}] = x^{1/2}\text{,}\) we find that
\begin{align*}
\int_0^1 (\sqrt{x} - x^2) \, dx =\mathstrut \amp \left. \frac{2}{3}x^{3/2} - \frac{1}{3}x^3 \right|_0^1\\
=\mathstrut \amp \frac{2}{3} - \frac{1}{3} - (0 - 0 )\\
=\mathstrut \amp \frac{1}{3}\text{.}
\end{align*}
