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