AP Calculus BC Quiz: Finding Antiderivatives And Indefinite Integrals

What this quiz covers

This quiz focuses on Finding Antiderivatives And Indefinite Integrals, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $8x^{3/2}-5x+C$ (correct answer)
2. $8x^{3/2}-5x$
3. $24x^{-1/2}-5+C$
4. $12x^{3/2}-5x+C$
5. $8x^{1/2}-5x+C$

Explanation: Finding antiderivatives involves applying basic integration rules to reverse the process of differentiation. To find the antiderivative of $F(x) = 12x^{1/2} - 5$, integrate each term by increasing the power by one and dividing by the new power. The integral of $12x^{1/2}$ is $8x^{3/2}$, and the integral of -5 is $-5x$, yielding $8x^{3/2} - 5x + C$. This reverses differentiation as the derivative of $8x^{3/2} - 5x + C$ is $12x^{1/2} - 5$. A tempting distractor like choice D fails because it uses 12 instead of 8 for the coefficient, forgetting to divide by $3/2$ properly. Always remember to add the constant of integration (C) and verify by differentiating the result.

Question 2

1. $\dfrac{9}{5}x^{5/3}-18x^{2/3}+C$ (correct answer)
2. $\dfrac{9}{5}x^{5/3}-18x^{2/3}$
3. $\dfrac{9}{5}x^{5/3}-36x^{2/3}+C$
4. $x^{5/3}-18x^{2/3}+C$
5. $2x^{-1/3}+4x^{-4/3}+C$

Explanation: This problem tests basic antiderivative reasoning skills. The antiderivative of $R'(x) = 3x^{2/3} - 12x^{-1/3}$ involves reversing differentiation with fractional powers. For $3x^{2/3}$, add 1 to get $5/3$, divide by $5/3$, resulting in $\dfrac{9}{5}x^{5/3}$. For $-12x^{-1/3}$, add 1 to get $2/3$, divide by $2/3$, yielding $-18x^{2/3}$. Choice C fails by doubling to $-36x^{2/3}$, mishandling the coefficient. Convert all terms to exponents and carefully compute new coefficients as a reliable strategy, adding $+C$.

Question 3

1. $3x^3+16x^{1/2}+C$ (correct answer)
2. $3x^3+4x^{1/2}+C$
3. $9x^3+16x^{1/2}+C$
4. $3x^3+16x^{1/2}$
5. $18x+4x^{-3/2}$

Explanation: This question requires finding an antiderivative by applying the power rule to each term separately. For C'(x) = 9x² + 8x⁻¹/², we integrate term by term: ∫9x² dx = 9x³/3 = 3x³, and ∫8x⁻¹/² dx = 8x¹/²/(1/2) = 16x¹/². Combining these results with the constant of integration gives 3x³ + 16x¹/² + C. The key insight is that x⁻¹/² integrates to x¹/² multiplied by 2, not divided by 2, because we divide by the new exponent (1/2). Choice B incorrectly computes the second term as 4x¹/², likely from misapplying the power rule by multiplying rather than dividing by 1/2. When integrating negative or fractional powers, always remember to add 1 to the exponent and divide by the new exponent.

Question 4

1. $\tan x+10x+C$ (correct answer)
2. $\tan x+10x$
3. $\sec^2 x+10+C$
4. $\sec x+10x+C$
5. $-\tan x+10x+C$

Explanation: This problem asks for an antiderivative of f'(x) = sec²x + 10. Finding antiderivatives reverses differentiation: the antiderivative of sec²x is tan x (since d/dx(tan x) = sec²x), and the antiderivative of 10 is 10x. Including the constant of integration gives tan x + 10x + C. Choice E shows -tan x + 10x + C, incorrectly using a negative sign with tan x. Remember: the derivative of tan x is sec²x, so the antiderivative of sec²x is tan x, not -tan x.

Question 5

1. $2t^3-2t^2+9t$
2. $2t^3-2t^2+9t+C$ (correct answer)
3. $12t-4+9$
4. $3t^2-4t+9+C$
5. $2t^3+2t^2+9t+C$

Explanation: This problem tests your ability to find antiderivatives using basic integration rules. To find an antiderivative of v(t) = 6t² - 4t + 9, we reverse the differentiation process by applying the power rule for integration: ∫tⁿ dt = tⁿ⁺¹/(n+1) + C. For 6t², we get 6t³/3 = 2t³; for -4t, we get -4t²/2 = -2t²; and for 9, we get 9t. Since antiderivatives are not unique, we must include the constant of integration C, giving us 2t³ - 2t² + 9t + C. Choice A is tempting because it shows the correct integration of each term, but it's missing the crucial constant C that represents the family of all antiderivatives. Remember that when finding antiderivatives, always add the constant of integration unless you're evaluating a definite integral.

Question 6

1. $t^{5}+3t^{-1}+C$ (correct answer)
2. $t^{5}-3t^{-1}+C$
3. $t^{5}+3t^{-1}$
4. $25t^{3}+6t^{-3}+C$
5. $5t^{5}+3t^{-1}+C$

Explanation: This problem tests basic antiderivative reasoning skills. The antiderivative of r(t) = 5t⁴ - 3t^{-2} is found by reversing differentiation using the power rule on each term. For 5t⁴, increase the exponent to 5 and divide by 5, giving t⁵. For -3t^{-2}, increase to -1 and divide by -1, resulting in 3t^{-1}. Choice B might tempt if one forgets to change the sign when integrating the negative term, but it incorrectly has -3t^{-1}. A key strategy for integrating power functions is to handle negative exponents carefully and always add +C.

Question 7

1. $t^7-\frac{1}{t^2}+C$ (correct answer)
2. $7t^7-\frac{1}{t^2}+C$
3. $t^7+\frac{1}{t^2}+C$
4. $t^7-\frac{1}{t^2}$
5. $42t^5-\frac{6}{t^4}$

Explanation: This problem involves integrating T'(t) = 7t⁶ + 2/t³, which we rewrite as T'(t) = 7t⁶ + 2t⁻³. Applying the power rule: ∫7t⁶ dt = 7t⁷/7 = t⁷, and ∫2t⁻³ dt = 2t⁻²/(-2) = -t⁻² = -1/t². The complete antiderivative is t⁷ - 1/t² + C. The positive term 2/t³ becomes negative after integration due to dividing by the negative new exponent -2. Choice C incorrectly shows +1/t², missing the sign change that occurs when integrating negative powers. When integrating t⁻ⁿ where n > 0, the result will have the opposite sign of the original coefficient due to division by a negative number.

Question 8

1. $4x-3x^3+\frac{1}{10}x^5+C$ (correct answer)
2. $4x-9x^3+\frac{1}{2}x^5+C$
3. $4-18x+2x^3+C$
4. $4x-3x^3+\frac{1}{10}x^5$
5. $4x-\frac{9}{2}x^3+\frac{1}{10}x^5+C$

Explanation: This problem tests integration of polynomial terms with various powers. To find an antiderivative of g(x) = 4 - 9x² + (1/2)x⁴, we integrate each term separately using the power rule. For the constant 4, we get 4x; for -9x², we get -9x³/3 = -3x³; for (1/2)x⁴, we get (1/2)x⁵/5 = x⁵/10. The complete antiderivative is 4x - 3x³ + x⁵/10 + C. Choice B (4x - 9x³ + (1/2)x⁵ + C) incorrectly integrates -9x² as -9x³ instead of -3x³, forgetting to divide by the new exponent 3. Always remember to divide by (n+1) when integrating xⁿ, as this step is essential for reversing the differentiation process.

Question 9

1. $3x^3+2\ln|x|+C$ (correct answer)
2. $3x^3+2\ln|x|$
3. $9x^3+2\ln|x|+C$
4. $3x^2+\dfrac{2}{x}+C$
5. $3x^3-2\ln|x|+C$

Explanation: This problem asks for an antiderivative of the marginal cost function $M(x) = 9x^2 + 2x^{-1}$. Finding antiderivatives means reversing differentiation: for each term, add 1 to the exponent and divide by the new exponent. For $9x^2$, we get $9x^3 / 3 = 3x^3$. For $2x^{-1}$, we add 1 to get $x^0 = 1$, but the antiderivative of $1/x$ is $\ln|x|$, so we get $2\ln|x|$. Including the constant of integration, the antiderivative is $3x^3 + 2\ln|x| + C$. Choice D incorrectly shows $3x^2 + \dfrac{2}{x} + C$, which appears to be a partial differentiation rather than integration. When integrating, remember that $\int (1/x) \, dx = \ln|x| + C$, not a power rule application.

Question 10

1. $3t^{4}+3t^{2}+C$ (correct answer)
2. $12t^{4}+6t^{2}+C$
3. $3t^{4}+3t^{2}$
4. $36t^{2}+6+C$
5. $3t^{4}-3t^{2}+C$

Explanation: This acceleration problem requires finding an antiderivative using the power rule for integration. To integrate $a(t) = 12t^3 + 6t$, we apply $\int t^n \, dt = \frac{t^{n+1}}{n+1} + C$ to each term: $12t^3$ becomes $\frac{12t^4}{4} = 3t^4$, and $6t$ becomes $\frac{6t^2}{2} = 3t^2$. The complete antiderivative is $3t^4 + 3t^2 + C$. Choice E shows $3t^4 - 3t^2 + C$, which has the wrong sign on the second term—this error occurs when students mistakenly change signs during integration, perhaps confusing it with differentiation rules. Remember that integration preserves the signs of the original terms unless the division step introduces a sign change (which only happens with negative exponents).

Question 11

1. $x^5+\frac{11}{2}x^2-9x+C$ (correct answer)
2. $x^5+11x^2-9x+C$
3. $5x^5+\frac{11}{2}x^2-9x+C$
4. $x^5+\frac{11}{2}x^2-9x$
5. $20x^3+11$

Explanation: To find an antiderivative of F(x) = 5x⁴ + 11x - 9, we integrate each term using the power rule. For 5x⁴: ∫5x⁴ dx = 5x⁵/5 = x⁵; for 11x: ∫11x dx = 11x²/2; and for -9: ∫(-9) dx = -9x. Combining these with the constant of integration gives x⁵ + (11/2)x² - 9x + C. The linear term 11x integrates to 11x²/2, not 11x² as shown in choice B, because we must divide by the new exponent 2. Choice D omits the constant C, which is essential for representing all antiderivatives. Always include '+ C' for indefinite integrals and carefully apply the division step in the power rule.

Question 12

1. $10t-3t^3+C$
2. $10t-t^3+C$ (correct answer)
3. $10-6t$
4. $10t-t^2+C$
5. $10t-t^3$

Explanation: Finding an antiderivative of acceleration gives velocity, which requires integrating a(t) = 10 - 3t². Applying the power rule term by term: ∫10 dt = 10t and ∫(-3t²) dt = -3t³/3 = -t³. Adding the constant of integration gives 10t - t³ + C. The coefficient -3 in the original function becomes -3/3 = -1 after integration, not -3 as shown in choice A. Choice A incorrectly keeps the coefficient as -3t³ instead of -t³, forgetting to divide by the new exponent. When integrating polynomial terms, always remember to divide the coefficient by (n+1) where n is the original exponent.

Question 13

1. $2t^3-2t^2+9t+C$ (correct answer)
2. $2t^3-2t^2+9t$
3. $12t-4+9+C$
4. $6t^3-4t^2+9t+C$
5. $2t^3-4t^2+9t+C$

Explanation: Finding antiderivatives involves applying basic integration rules to reverse the process of differentiation. To find the antiderivative of $v(t) = 6t^2 - 4t + 9$, integrate each term separately by increasing the exponent by one and dividing by the new exponent. The integral of $6t^2$ is $2t^3$, the integral of $-4t$ is $-2t^2$, and the integral of 9 is $9t$, resulting in $2t^3 - 2t^2 + 9t + C$. This reverses differentiation because differentiating $2t^3 - 2t^2 + 9t + C$ yields back $6t^2 - 4t + 9$. A tempting distractor like choice E fails because it uses $-4t^2$ instead of $-2t^2$, which comes from incorrectly integrating $-4t$ without dividing by 2. Always remember to add the constant of integration (C) and verify by differentiating the result.

Question 14

1. $2\sin x+5x+C$ (correct answer)
2. $2\sin x+5x$
3. $2\cos x+5+C$
4. $-2\sin x+5x+C$
5. $2\sin x+5+C$

Explanation: Finding antiderivatives involves applying basic integration rules to reverse the process of differentiation. To find the antiderivative of (f'(x) = 2cos x + 5), recall that the integral of (cos x) is (sin x) and of a constant is the variable times that constant. The integral is (2sin x + 5x + C). This reverses differentiation since the derivative of (2sin x + 5x + C) is (2cos x + 5). A tempting distractor like choice D fails because it uses (-2sin x) instead of (2sin x), confusing the integral of cosine with negative sine. Always remember to add the constant of integration (C) and verify by differentiating the result.

Question 15

1. $-7t^{-1}+4t+C$ (correct answer)
2. $7t^{-1}+4t+C$
3. $-7t^{-1}+4t$
4. $-14t^{-1}+4t+C$
5. $7t^{-2}+4+C$

Explanation: Finding antiderivatives involves applying basic integration rules to reverse the process of differentiation. To find the antiderivative of (T'(t) = 7t^{-2} + 4), integrate term by term by adding one to each exponent and dividing by the new exponent. The integral of (7t^{-2}) is (-7t^{-1}), and the integral of 4 is (4t), giving (-7t^{-1} + 4t + C). This reverses differentiation since differentiating (-7t^{-1} + 4t + C) returns (7t^{-2} + 4). A tempting distractor like choice B fails because it omits the negative sign from integrating (t^{-2}), which should produce a -1 in the denominator. Always remember to add the constant of integration (C) and verify by differentiating the result.

Question 16

1. $8t+t^{-2}+C$ (correct answer)
2. $8t-t^{-2}+C$
3. $8t+t^{-2}$
4. $8-2t^{-3}+C$
5. $8t+2t^{-2}+C$

Explanation: Finding antiderivatives involves applying basic integration rules to reverse the process of differentiation. To find the antiderivative of (r(t) = 8 - 2t^{-3}), integrate each term by adding one to the exponent and dividing by the new exponent. The integral of 8 is (8t), and the integral of (-2t^{-3}) is (t^{-2}), resulting in (8t + t^{-2} + C). This reverses differentiation because differentiating (8t + t^{-2} + C) gives back (8 - 2t^{-3}). A tempting distractor like choice B fails because it incorrectly assigns a negative sign to (t^{-2}), mishandling the integration of the negative coefficient. Always remember to add the constant of integration (C) and verify by differentiating the result.

Question 17

1. $2t^3-2t^2+5t$
2. $2t^3-2t^2+5t+C$ (correct answer)
3. $12t-4+5$
4. $6t^3-4t^2+5t+C$
5. $2t^3-4t^2+5t+C$

Explanation: This problem tests your ability to find antiderivatives using basic integration rules. To find an antiderivative of v(t) = 6t² - 4t + 5, we reverse the differentiation process by applying the power rule for integration: ∫tⁿ dt = tⁿ⁺¹/(n+1) + C. For 6t², we get 6t³/3 = 2t³; for -4t, we get -4t²/2 = -2t²; and for 5, we get 5t. Since antiderivatives are not unique, we must include the constant of integration C, giving us 2t³ - 2t² + 5t + C. Choice A incorrectly omits the constant C, which is a common error when students forget that indefinite integrals represent families of functions. Remember that every antiderivative must include '+ C' to account for all possible vertical shifts of the function.

Question 18

1. $7e^t-\dfrac{4}{t}+C$
2. $7e^t+\dfrac{4}{t}+C$ (correct answer)
3. $7e^t+\dfrac{4}{t}$
4. $7e^t-\dfrac{4}{t^2}+C$
5. $7e^t+\dfrac{4}{t^2}+C$

Explanation: This problem requires finding an antiderivative of a(t) = 7eᵗ - 4/t². To find antiderivatives, we reverse differentiation for each term. The antiderivative of eᵗ is eᵗ (since d/dt(eᵗ) = eᵗ), so 7eᵗ integrates to 7eᵗ. For -4/t² = -4t⁻², we add 1 to get t⁻¹, and dividing by -1 gives us 4/t. Including the constant of integration yields 7eᵗ + 4/t + C. Choice A incorrectly shows -4/t instead of +4/t, missing the sign change when integrating negative powers. Remember: when integrating t⁻ⁿ where n ≠ 1, the result is t⁻ⁿ⁺¹/(-n+1), and pay attention to signs.

Question 19

1. $4x-5x^{2}+C$ (correct answer)
2. $4x-10x^{2}+C$
3. $4-10x+C$
4. $4x-5x^{2}$
5. $4x+5x^{2}+C$

Explanation: This problem tests basic antiderivative skills with linear functions. To find an antiderivative of F(x) = 4 - 10x, we integrate each term: the constant 4 becomes 4x (since ∫c dx = cx), and -10x becomes -10x²/2 = -5x². Adding the constant of integration gives us 4x - 5x² + C. Choice B incorrectly shows 4x - 10x² + C, which results from forgetting to divide by the new exponent when integrating -10x—a common mistake where students only increase the exponent without completing the power rule formula. The key strategy for polynomial integration is to systematically apply xⁿ⁺¹/(n+1) to each term, treating constants as x⁰ terms.

Question 20

1. $\dfrac{1}{12}x^{4}+6x^{-1}+C$ (correct answer)
2. $\dfrac{1}{12}x^{4}-6x^{-1}+C$
3. $\dfrac{1}{9}x^{4}+6x^{-1}+C$
4. $\dfrac{1}{12}x^{4}+6x^{-1}$
5. $x^{2}+12x^{-3}+C$

Explanation: This problem tests basic antiderivative reasoning skills. To find the antiderivative of F(x) = (1/3)x³ - 6x^{-2}, reverse differentiation term by term. For (1/3)x³, add one to get x⁴ and divide by 4, giving (1/12)x⁴. For -6x^{-2}, add one to get x^{-1} and divide by -1, resulting in 6x^{-1}. Choice B tempts by negating to -6x^{-1}, but the signs reverse correctly to positive. A transferable approach is to integrate each term independently, simplify, and verify by differentiation while including +C.