Antiderivatives & Indefinite Integrals - Calculus 1

1. The indefinite integral $\int \frac{4}{x^2+1} dx$ evaluates to which family of functions?
   Answer: $4\arctan(x) + C$ This integral matches the form of the derivative of the inverse tangent function. The constant 4 can be factored out: $\int \frac{4}{x^2+1} dx = 4 \int \frac{1}{1+x^2} dx$ The integral $\int \frac{1}{1+x^2} dx$ is a standard result: $\int \frac{1}{1+x^2} dx = \arctan(x) + C$ Therefore, the original integral is $4\arctan(x) + C$.
2. Let $F(x)$ be a function such that $F''(x) = 6x + \cos(x)$. Which of the following represents the most general form of $F(x)$?
   Answer: $x^3 - \cos(x) + C_1x + C_2$ To find $F(x)$ from $F''(x)$, we must integrate twice. First, find $F'(x)$ by integrating $F''(x)$: $F'(x) = \int (6x + \cos(x)) dx = 6\frac{x^2}{2} + \sin(x) + C_1 = 3x^2 + \sin(x) + C_1$ Now, find $F(x)$ by integrating $F'(x)$: $F(x) = \int (3x^2 + \sin(x) + C_1) dx = 3\frac{x^3}{3} - \cos(x) + C_1x + C_2$ Simplifying gives the most general form of $F(x)$: $F(x) = x^3 - \cos(x) + C_1x + C_2$ where $C_1$ and $C_2$ are arbitrary constants.
3. Let $F(x)$ be the most general antiderivative of $f(x) = 3 \cdot 5^x$. Which of the following is $F(x)$?
   Answer: $3 \frac{5^x}{\ln 5} + C$ To find the antiderivative of $f(x) = 3 \cdot 5^x$, we use the constant multiple rule and the rule for integrating exponential functions of the form $a^x$. The rule is $\int a^x dx = \frac{a^x}{\ln a} + C$. $\int 3 \cdot 5^x dx = 3 \int 5^x dx = 3 \left(\frac{5^x}{\ln 5}\right) + C = \frac{3 \cdot 5^x}{\ln 5} + C$
4. Which of the following is an antiderivative of $f(x) = \frac{2x^2 - 3}{x^2 + 1}$?
   Answer: $2x - 5\arctan(x)$ The integrand is an improper rational function. We can rewrite it using algebraic manipulation. $f(x) = \frac{2x^2 - 3}{x^2 + 1} = \frac{2(x^2 + 1) - 2 - 3}{x^2 + 1} = \frac{2(x^2 + 1) - 5}{x^2 + 1} = \frac{2(x^2+1)}{x^2+1} - \frac{5}{x^2+1} = 2 - \frac{5}{x^2+1}$. Now, we integrate this simplified form: $\int (2 - \frac{5}{x^2+1}) dx = \int 2 dx - 5\int \frac{1}{x^2+1} dx$. This evaluates to $2x - 5\arctan(x) + C$. Any function of the form $2x - 5\arctan(x) + C$ is an antiderivative.
5. Evaluate the indefinite integral $\int (3^x - e^3) dx$.
   Answer: $\frac{3^x}{\ln(3)} - e^3 x + C$ The integral can be split into two parts: $\int 3^x dx - \int e^3 dx$. The first part is the integral of an exponential function with a base other than $e$, for which the rule is $\int a^x dx = \frac{a^x}{\ln(a)} + C$. So, $\int 3^x dx = \frac{3^x}{\ln(3)}$. The second part, $\int e^3 dx$, is the integral of a constant, because $e^3$ is just a number. The integral of a constant $k$ is $kx$. So, $\int e^3 dx = e^3 x$. Combining these results, we get $\frac{3^x}{\ln(3)} - e^3 x + C$.
6. An object's velocity is given by $v(t) = \cos(2t)$. Find the most general function for the object's position, $s(t)$, given that $s'(t) = v(t)$.
   Answer: $\frac{1}{2}\sin(2t) + C$ To find the position function $s(t)$, we must find the antiderivative of the velocity function $v(t)$: $s(t) = \int \cos(2t) dt$ This requires a reversal of the chain rule. The integral of $\cos(u)$ is $\sin(u)$. If we guess $\sin(2t)$, its derivative is $\cos(2t) \cdot 2$. Since our integrand is missing the factor of 2, we must compensate by multiplying by $\frac{1}{2}$. Therefore: $s(t) = \frac{1}{2}\sin(2t) + C$
7. If $f(x) = x^2 \cos(x)$, which of the following is $\int f'(x) dx$?
   Answer: $x^2 \cos(x) + C$ This question tests the fundamental relationship between differentiation and integration. The indefinite integral of the derivative of a function, $\int f'(x) dx$, is the original function $f(x)$ plus a constant of integration, $C$. This is a direct consequence of the Fundamental Theorem of Calculus. Therefore, $\int f'(x) dx = f(x) + C$. Given $f(x) = x^2 \cos(x)$, the result is $x^2 \cos(x) + C$.
8. Let $f(x)$ be a function defined as $f(x) = \begin{cases} 2x + 1 & \text{if } x < 1 \\ 3x^2 & \text{if } x \ge 1 \end{cases}$. If $F(x)$ is a continuous antiderivative of $f(x)$ such that $F(2) = 10$, find $F(0)$.
   Answer: 1 We find the antiderivative for each piece. For $x < 1$, $F(x) = \int (2x+1) dx = x^2 + x + C_1$. For $x \ge 1$, $F(x) = \int (3x^2) dx = x^3 + C_2$. We use the condition $F(2)=10$ with the second piece, since $2 \ge 1$: $F(2) = 2^3 + C_2 = 8 + C_2 = 10$, which implies $C_2 = 2$. So, for $x \ge 1$, $F(x) = x^3 + 2$. Since $F(x)$ must be continuous at $x=1$, the values of the two pieces must be equal at $x=1$. So, $1^2 + 1 + C_1 = 1^3 + 2$, which simplifies to $2 + C_1 = 3$, giving $C_1 = 1$. Thus, for $x < 1$, $F(x) = x^2 + x + 1$. We need to find $F(0)$. Since $0 < 1$, we use the first piece: $F(0) = 0^2 + 0 + 1 = 1$.
9. Evaluate $\int (2x - \frac{1}{\sqrt{x}})^2 dx$ for $x>0$.
   Answer: $\frac{4}{3}x^3 - \frac{8}{3}x^{3/2} + \ln(x) + C$ To evaluate this integral, we first expand the integrand. Let $x^{-1/2} = \frac{1}{\sqrt{x}}$. $(2x - x^{-1/2})^2 = (2x)^2 - 2(2x)(x^{-1/2}) + (x^{-1/2})^2 = 4x^2 - 4x^{1/2} + x^{-1}$. Now we can integrate term by term: $\int (4x^2 - 4x^{1/2} + x^{-1}) dx = 4\int x^2 dx - 4\int x^{1/2} dx + \int x^{-1} dx$. Applying the power rule to the first two terms and the log rule to the third term gives $4(\frac{x^3}{3}) - 4(\frac{x^{3/2}}{3/2}) + \ln|x| + C$. Simplifying this expression yields $\frac{4}{3}x^3 - \frac{8}{3}x^{3/2} + \ln(x) + C$ (since $x>0$, $|x|=x$).
10. If $F(x)$ and $G(x)$ are two different functions such that $F'(x) = G'(x)$ for all $x$ in $(-\infty, \infty)$, which of the following must be true?
    Answer: The graphs of $F(x)$ and $G(x)$ are vertical translations of each other. If two functions $F(x)$ and $G(x)$ have the same derivative on an interval, then they must differ by a constant. Let $H(x) = F(x) - G(x)$. Then $H'(x) = F'(x) - G'(x) = 0$. A function whose derivative is zero everywhere must be a constant function, so $H(x) = C$ for some constant $C$. This means $F(x) - G(x) = C$, or $F(x) = G(x) + C$. Graphically, adding a constant $C$ to a function shifts its graph vertically. Therefore, the graphs of $F(x)$ and $G(x)$ are vertical translations of each other.
11. Let $F(x)$ be an antiderivative of a continuous function $f(x)$. Which of the following is an antiderivative of the function $g(x) = f(x) + 2$?
    Answer: $F(x) + 2x$ To find an antiderivative of $g(x) = f(x) + 2$, we integrate $g(x)$ with respect to $x$: $\int g(x) dx = \int (f(x) + 2) dx$ Using the sum rule for integrals: $\int f(x) dx + \int 2 dx$ We are given that $F(x)$ is an antiderivative of $f(x)$, so $\int f(x) dx = F(x) + C$. The integral of 2 is $2x$. Combining these, the most general antiderivative is $F(x) + 2x + C$. Therefore, $F(x) + 2x$ is an antiderivative of $g(x)$.
12. Evaluate $\int \frac{x+1}{x} dx$.
    Answer: $x + \ln|x| + C$ To solve this integral, first simplify the integrand by splitting the fraction: $\int \frac{x+1}{x} dx = \int \left(\frac{x}{x} + \frac{1}{x}\right) dx = \int \left(1 + \frac{1}{x}\right) dx$ Now, integrate term by term: $\int 1 dx + \int \frac{1}{x} dx = x + \ln|x| + C$ This is a common problem where algebraic simplification is required before integration.
13. Evaluate $\int (3e^t - \sec^2(t)) dt$.
    Answer: $3e^t - \tan(t) + C$ We integrate term by term using standard integration rules. $\int (3e^t - \sec^2(t)) dt = \int 3e^t dt - \int \sec^2(t) dt$ For the first term, the integral of $e^t$ is $e^t$. For the second term, the integral of $\sec^2(t)$ is $\tan(t)$. $3\int e^t dt - \int \sec^2(t) dt = 3e^t - \tan(t) + C$
14. Let $g(x) = \frac{x^4}{4} - \cos(x)$. Which of the following represents $\int g'(x) dx$?
    Answer: $\frac{x^4}{4} - \cos(x) + C$ This question tests the relationship between a function, its derivative, and its antiderivative. The expression $\int g'(x) dx$ asks for the antiderivative of the derivative of $g(x)$. According to the Fundamental Theorem of Calculus, integrating the derivative of a function returns the original function, plus an arbitrary constant of integration. Therefore, $\int g'(x) dx = g(x) + C$. Substituting the given expression for $g(x)$ yields $\frac{x^4}{4} - \cos(x) + C$.
15. Find the family of functions whose derivative is $h'(x) = \frac{x^3 - 5}{x^3}$.
    Answer: $x + \frac{5}{2}x^{-2} + C$ To find the family of functions $h(x)$, we need to compute the indefinite integral of $h'(x)$. First, simplify the integrand: $\int \frac{x^3 - 5}{x^3} dx = \int \left(\frac{x^3}{x^3} - \frac{5}{x^3}\right) dx = \int (1 - 5x^{-3}) dx$ Now, integrate term-by-term using the power rule: $\int 1 dx - 5\int x^{-3} dx = x - 5\left(\frac{x^{-3+1}}{-3+1}\right) + C = x - 5\left(\frac{x^{-2}}{-2}\right) + C = x + \frac{5}{2}x^{-2} + C$
16. Which of the following expressions represents the family of all functions whose slope of the tangent line at any point $x$ is given by $m(x) = x(x-2)$?
    Answer: $\frac{1}{3}x^3 - x^2 + C$ The slope of the tangent line to a function at a point $x$ is given by its derivative, $f'(x)$. So, we are given $f'(x) = m(x) = x(x-2) = x^2 - 2x$. To find the family of all such functions, we must find the indefinite integral of $f'(x)$. $$f(x) = \int (x^2 - 2x) dx$$ Using the power rule for integration: $$f(x) = \frac{x^3}{3} - 2\frac{x^2}{2} + C = \frac{1}{3}x^3 - x^2 + C$$
17. Find the most general antiderivative of $f(\theta) = 1 + \tan^2\theta$.
    Answer: $\tan\theta + C$ This problem requires knowledge of trigonometric identities. Recall the Pythagorean identity: $\sec^2\theta = 1 + \tan^2\theta$. By substituting this into the integral, we get: $$\int (1 + \tan^2\theta) d\theta = \int \sec^2\theta d\theta$$ The integral of $\sec^2\theta$ is a standard result, as the derivative of $\tan\theta$ is $\sec^2\theta$. Therefore: $$\int \sec^2\theta d\theta = \tan\theta + C$$
18. Let $F(x)$ be the antiderivative of $f(x) = 3x^2 - 2\sec^2(x)$. If the graph of $F(x)$ passes through the point $(\pi, -2\pi)$, which of the following is the expression for $F(x)$?
    Answer: $x^3 - 2\tan(x) - \pi^3 - 2\pi$ To find $F(x)$, we must compute the indefinite integral of $f(x)$: $F(x) = \int (3x^2 - 2\sec^2(x)) dx = x^3 - 2\tan(x) + C$. We are given that $F(\pi) = -2\pi$. We use this condition to solve for the constant of integration, $C$. $F(\pi) = (\pi)^3 - 2\tan(\pi) + C = -2\pi$. Since $\tan(\pi) = 0$, the equation becomes $\pi^3 - 0 + C = -2\pi$, which gives $C = -2\pi - \pi^3$. Substituting this value of $C$ back into the expression for $F(x)$ yields $F(x) = x^3 - 2\tan(x) - \pi^3 - 2\pi$.
19. Find the indefinite integral $\int (e^2 + \pi) dx$.
    Answer: $e^2 x + \pi x + C$ The integrand, $e^2 + \pi$, is a constant because $e$ and $\pi$ are constant numbers. Let $k = e^2 + \pi$. The integral is $\int k dx$. The integral of a constant $k$ with respect to $x$ is $kx + C$. Therefore: $$\int (e^2 + \pi) dx = (e^2 + \pi)x + C = e^2 x + \pi x + C$$ A common mistake is to treat $e$ as a variable and apply the power rule.
20. Which of the following is the most general antiderivative of $f(x) = \frac{5}{\sqrt{1-x^2}}$?
    Answer: $5\arcsin(x) + C$ The integral is $\int \frac{5}{\sqrt{1-x^2}} dx$. We can factor out the constant 5: $5 \int \frac{1}{\sqrt{1-x^2}} dx$ The integral $\int \frac{1}{\sqrt{1-x^2}} dx$ is the standard form for the arcsine function. Therefore: $5 \int \frac{1}{\sqrt{1-x^2}} dx = 5\arcsin(x) + C$

Start this quiz