Section 4.9: Antiderivatives
1. Evaluate each integral. (This is same as finding the most general antiderivative of the function inside the integral sign) solution
(a) \( \int x^3 \, dx\) (b) \( \int e^x \, dx\) (c) \( \int \frac{1}{x}\,
dx\) (d) \( \int \sqrt x \, dx\) (e) \( \int \sqrt[3] {x^2} \, dx\) (f) \( \int \, dx\) (g) \( \int 0 \, dx\) (h) \( \int 2 \, dx\) (i) \( \int \pi^2 \, dx\) (j) \( \int e^3 \, dx\) (k) \(\int x^{99} \, dx\) (l) \( \int \frac{1}{x^4} \, dx\)
(m) \( \int 3^x dx\) (n) \(\int 5 \cdot 2^x \, dx\)
2. Find the most general antiderivative of the function. solution
(a) \(f(x)= 1-3x^2+e^x \) (b) \(\displaystyle f(x)=\sqrt[4]{x} - \frac{1}{x^2}\) (c) \( f(x)= \dfrac{1+x-x^2}{x}\)
3. Find the most general antiderivative of the function. solution
(a) \(f(t)= 2t(2-t)^2 \) (b) \(f(x)=\dfrac{x^2+2x-4}{\sqrt x}\) (c) \( f(x)= (2x-1)(x+3)\)
4. Find the most general antiderivative of the function. solution
(a) \( f(\theta)=\pi -\sec \theta \tan \theta \) (b) \( f(x) = e^x+\cos x + \dfrac{1}{\cos^2 x} \) (c) \( f(x)= 14\,\sqrt[4]{x^3} + \dfrac{2}{1+x^2} \)
5. Find \(f(2)\) given that the graph of \(f\) passes through the point \( (1, 6)\) and that the slope of its tangent line at \((x, f(x))\) is
\(2x + 1\). solution
6. Find the function \( f \) in each case.
(a) If \( f'(x)=1-6x \), and \(f(1)=8\),
find \(f\).
solution
(b) If \( f''(x)=\sin x + \cos x\), and \(f(0)=3,
f'(0)=4 \), find \(f\).
solution
(c) If \( f''(t)= 2 e^t + 3 \sin t\), and \(f(0)=0, f(\pi)=0 \),
find \(f\).
solution
7. A particle is moving with the following data. Find the position \(s(t)\) of the particle. solution
\[ v(t)= 3 \sqrt{t}+1, \quad s(4)=10 \]
8. Solve the initial value problem.
(a) \( f'(x) = 1+3x^2, \quad f(-1) = 2 \)
solution
(b) \( f'(x) = 2 x - \dfrac{1}{x}, \quad f(1) = 5 \)
solution
(c) \( f'(x) = \dfrac{1}{\sqrt x}, \quad f(4) = 2 \)
solution
Evaluate \( \displaystyle{\int \left(\sin(3x) - e^{5x} \right) \, dx}\)
solution