Finding Antiderivatives and Indefinite Integrals - AP Calculus AB

The derivative is $m'(x)=4x^3+9x^8$ . Which is an antiderivative of $m'(x)$ ?
Answer: $x^4+x^9+C$ This question tests the skill of finding antiderivatives by reversing the power rule for differentiation. To find the antiderivative of m'(x) = 4x³ + 9x⁸, increase the exponent of each term by one and divide by the new exponent, resulting in (4/4)x⁴ + (9/9)x⁹ + C. This process reverses differentiation because differentiating x⁴ + x⁹ + C yields 4x³ + 9x⁸, matching the original derivative. The constant C accounts for the fact that differentiation eliminates constants, so antiderivatives differ by a constant. A tempting distractor like choice C fails because it incorrectly uses coefficients of 4 and 9 without dividing properly. Always remember that integration is the reverse of differentiation, so check your antiderivative by differentiating it back to see if you get the original function.
The rate of change of a quantity is $Q'(t)=4t^{7}$ . Which function is an antiderivative of $Q'(t)$ ?
Answer: $\frac{1}{2}t^{8}+C$ Finding an antiderivative means reversing the differentiation process using integration rules. For Q'(t) = 4t⁷, we apply the power rule for integration: ∫4t⁷ dt = 4·t⁸/8 + C = t⁸/2 + C. The antiderivative increases the exponent from 7 to 8 and divides by the new exponent, simplifying 4/8 to 1/2. Choice D (1/2·t⁸) might look correct but is missing the crucial constant of integration C. Choice E (28t⁶) represents the derivative of 4t⁷, not its antiderivative—a common confusion when students mix up differentiation and integration. Remember: to find an antiderivative, add 1 to the exponent, divide by the new exponent, and always include the constant C.
A population grows at rate $P'(t)=12t^5-5t^2$ . Find an antiderivative of $P'(t)$ .
Answer: $2t^6-\frac{5}{3}t^3+C$ This problem requires finding an antiderivative of the population growth rate P'(t) = 12t⁵ - 5t², demonstrating basic antiderivative reasoning. To find the antiderivative, we reverse the differentiation process using the power rule for integration. For 12t⁵, we get 12t⁶/6 = 2t⁶; for -5t², we get -5t³/3 = -5/3·t³. Combining these with the constant of integration gives 2t⁶ - (5/3)t³ + C. Choice B (2t⁶ - 5t³ + C) incorrectly integrates -5t² as -5t³ instead of -(5/3)t³, forgetting to divide by the new exponent. The key integration strategy is to carefully apply the power rule formula ∫tⁿdt = $t^{n+1}$ /(n+1) + C, ensuring you divide by the new exponent (n+1).
The marginal cost is $M(x)=9x^2+2x$ . Which is an antiderivative of $M(x)$ ?
Answer: $3x^3+x^2+C$ This question tests the skill of finding antiderivatives by reversing the power rule for differentiation. To find the antiderivative of M(x) = 9x² + 2x, increase the exponent of each term by one and divide by the new exponent, resulting in (9/3)x³ + (2/2)x² + C. This process reverses differentiation because differentiating 3x³ + x² + C yields 9x² + 2x, matching the original marginal cost function. The constant C accounts for the fact that differentiation eliminates constants, so antiderivatives differ by a constant. A tempting distractor like choice D fails because it incorrectly uses coefficients of 9 and 2 instead of dividing properly. Always remember that integration is the reverse of differentiation, so check your antiderivative by differentiating it back to see if you get the original function.
The marginal cost is $C'(x)=0.2x^4-3x$ . Which is an antiderivative of $C'(x)$ ?
Answer: $0.04x^5-$ \frac{3}{2}x^2+C$ Finding the antiderivative involves basic reasoning about reversing the power rule for differentiation. To find the antiderivative of $0.2x^4$ , increase the exponent to 5 and divide by 5, yielding 0.2(1/5) $x^5$ = $0.04x^5$ . For -3x, increase to 2 and divide by 2, resulting in -(3/2) $x^2$ . Combine these terms and add the constant C. A tempting distractor is choice C, which converts decimals incorrectly. To master antiderivatives, always increase the exponent by one, divide by the new exponent, and include the constant C for indefinite integrals.
A function has derivative $g'(x)=3x^8-10x^4+2$ . Which is a general antiderivative of $g'(x)$ ?
Answer: $\frac{1}{3}x^9-2x^5+2x+C$ This problem asks for a general antiderivative of g'(x) = 3x⁸ - 10x⁴ + 2, testing basic antiderivative reasoning skills. To find the antiderivative, we apply the power rule for integration to each term. For 3x⁸, we get 3x⁹/9 = (1/3)x⁹; for -10x⁴, we get -10x⁵/5 = -2x⁵; for the constant 2, we get 2x. Combining these with the constant of integration yields (1/3)x⁹ - 2x⁵ + 2x + C. Choice D ((1/9)x⁹ - 2x⁵ + 2x + C) incorrectly integrates 3x⁸ as (1/9)x⁹ instead of (1/3)x⁹, making an error in simplifying 3/9. The transferable strategy is to carefully simplify coefficients after applying the power rule: when integrating $ax^n$ , the result is a/(n+1)· $x^{n+1}$ .
The derivative is $g'(t)=5t^{2/3}$ . Which is an antiderivative of $g'(t)$ ?
Answer: $3t^{5/3}+C$ Finding the antiderivative involves basic reasoning about reversing the power rule for differentiation. To find the antiderivative of $5t^{2/3}$ , increase the exponent to $5/3$ and divide by $5/3$ , yielding $5(3/5)t^{5/3} = 3t^{5/3}$ . There are no other terms, so add the constant C. This handles fractional exponents correctly. A tempting distractor is choice B, which forgets to divide properly. To master antiderivatives, always increase the exponent by one, divide by the new exponent, and include the constant C for indefinite integrals.
The marginal cost is $C'(x)=12x^{3}+6x$ . Which expression is an antiderivative of $C'(x)$ ?
Answer: $3x^{4}+3x^{2}+C$ Finding an antiderivative requires applying the power rule for integration to reverse differentiation. For C'(x) = 12x³ + 6x, we integrate each term: the antiderivative of 12x³ is 12x⁴/4 = 3x⁴, and the antiderivative of 6x is 6x²/2 = 3x². The complete antiderivative is 3x⁴ + 3x² + C. Choice E (3x⁴ + 6x² + C) is incorrect because it fails to divide the coefficient 6 by 2 when integrating 6x—a common error when students forget to divide by the new exponent. The key strategy is to increase each exponent by 1, divide the coefficient by the new exponent, and always include the constant of integration C.
A function satisfies $g'(t)=18t^2+5t^4$ . Which is an antiderivative of $g'(t)$ ?
Answer: $6t^3+t^5+C$ Finding the antiderivative involves basic reasoning about reversing the power rule for differentiation. To find the antiderivative of $18t^2$ , increase the exponent to 3 and divide by 3, yielding 18(1/3) $t^3$ = $6t^3$ . For $5t^4$ , increase to 5 and divide by 5, resulting in $t^5$ . Combine these terms and add the constant C. A tempting distractor is choice C, which multiplies by the original coefficients without proper division. To master antiderivatives, always increase the exponent by one, divide by the new exponent, and include the constant C for indefinite integrals.
The rate of water flow is $r(t)=5t^4+8$ . Which function is an antiderivative of $r(t)$ ?
Answer: $t^5+8t+C$ This question tests the skill of finding antiderivatives by reversing the power rule for differentiation. To find the antiderivative of r(t) = 5t⁴ + 8, increase the exponent of each term by one and divide by the new exponent, resulting in (5/5)t⁵ + 8t + C. This process reverses differentiation because differentiating t⁵ + 8t + C yields 5t⁴ + 8, matching the original rate function. The constant C accounts for the fact that differentiation eliminates constants, so antiderivatives differ by a constant. A tempting distractor like choice D fails because it incorrectly uses a coefficient of 5 instead of 1 for the t⁵ term. Always remember that integration is the reverse of differentiation, so check your antiderivative by differentiating it back to see if you get the original function.
The slope of a curve is $\frac{dy}{dx}=15x^4-2x^3$ . Which $y$ is an antiderivative?
Answer: $3x^5-\frac{1}{2}x^4+C$ This problem asks for an antiderivative y given that dy/dx = 15x⁴ - 2x³, testing basic antiderivative reasoning skills. To find y, we integrate the derivative using the power rule for integration. For 15x⁴, we get 15x⁵/5 = 3x⁵; for -2x³, we get -2x⁴/4 = -(1/2)x⁴. Including the constant of integration yields y = 3x⁵ - (1/2)x⁴ + C. Choice B (3x⁵ - 2x⁴ + C) incorrectly integrates -2x³ as -2x⁴ instead of -(1/2)x⁴, forgetting to divide by the new exponent 4. The transferable integration-differentiation reversal strategy is to always divide by (n+1) when integrating xⁿ, which reverses the multiplication by n that occurs during differentiation.
A particle's velocity is $v(t)=6t^2-4t+9$ . Which function is an antiderivative of $v(t)$ ?
Answer: $2t^3-2t^2+9t+C$ This problem requires finding an antiderivative of the velocity function v(t) = 6t² - 4t + 9, which tests basic antiderivative reasoning skills. To find an antiderivative, we reverse the differentiation process by applying the power rule in reverse: for each term tⁿ, the antiderivative is $t^{n+1}$ /(n+1). For 6t², we get 6t³/3 = 2t³; for -4t, we get -4t²/2 = -2t²; and for the constant 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 (2t³ - 2t² + 9t) is tempting but incorrect because it omits the essential constant of integration that represents the family of all antiderivatives. The key strategy is to remember that indefinite integrals always include +C to represent all possible vertical shifts of the antiderivative function.
The acceleration of a cart is $a(t)=8t^{3}+5$ . Which function is an antiderivative of $a(t)$ ?
Answer: $2t^{4}+5t+C$ Finding an antiderivative requires reversing the differentiation process using integration rules. For a(t) = 8t³ + 5, we apply the power rule: the antiderivative of 8t³ is 8t⁴/4 = 2t⁴, and the antiderivative of 5 is 5t. Therefore, the complete antiderivative is 2t⁴ + 5t + C, where C is the constant of integration. Choice B (2t⁴ + 5t) is incorrect because it omits the constant C, which is essential for representing all possible antiderivatives that differ by a constant. When integrating, remember to add 1 to the exponent and divide by the new exponent, and always include the constant C unless working with definite integrals.
A runner's velocity is $v(t)=\dfrac{3}{2}t^4-5t$ . Which is an antiderivative of $v(t)$ ?
Answer: $\dfrac{3}{10}t^5-\dfrac{5}{2}t^2+C$ Finding antiderivatives requires careful application of the power rule, especially with fractional coefficients. To integrate v(t) = (3/2)t⁴ - 5t, we apply the rule to each term: (3/2)t⁴ becomes (3/2)t⁵/5 = (3/10)t⁵, and -5t becomes -5t²/2 = -(5/2)t². The antiderivative is (3/10)t⁵ - (5/2)t² + C. Choice E ((3/2)t⁵ - (5/2)t² + C) incorrectly keeps the original coefficient 3/2 when integrating (3/2)t⁴, forgetting to divide by the new exponent 5. When integrating terms with fractions, multiply the coefficient by 1/(n+1), which often creates more complex fractions that must be simplified.
The velocity of a runner is $v(t)=15t^2+6t-1$ . Which is an antiderivative of $v(t)$ ?
Answer: $5t^3+3t^2-t+C$ This question tests the skill of finding antiderivatives by reversing the power rule for differentiation. To find the antiderivative of v(t) = 15t² + 6t - 1, increase the exponent of each term by one and divide by the new exponent, resulting in (15/3)t³ + (6/2)t² - t + C. This process reverses differentiation because differentiating 5t³ + 3t² - t + C yields 15t² + 6t - 1, matching the original velocity. The constant C accounts for the fact that differentiation eliminates constants, so antiderivatives differ by a constant. A tempting distractor like choice D fails because it incorrectly uses a coefficient of 15 instead of 5 for the t³ term. Always remember that integration is the reverse of differentiation, so check your antiderivative by differentiating it back to see if you get the original function.
A curve has slope $y'(x)=25x^4$ . Which is an antiderivative of $y'(x)$ ?
Answer: $5x^5+C$ Finding the antiderivative involves basic reasoning about reversing the power rule for differentiation. To find the antiderivative of $25x^4$ , increase the exponent to 5 and divide by 5, yielding $5x^5$ . There are no other terms, so add the constant C. This reverses the differentiation process directly. A tempting distractor is choice B, which multiplies by 5 again incorrectly. To master antiderivatives, always increase the exponent by one, divide by the new exponent, and include the constant C for indefinite integrals.
The marginal revenue is $R'(x)=4x^9-\dfrac{1}{3}x^2$ . Which is an antiderivative of $R'(x)$ ?
Answer: $\dfrac{2}{5}x^{10}-\dfrac{1}{9}x^3+C$ Finding antiderivatives with fractional coefficients requires careful algebraic manipulation during integration. For R'(x) = 4x⁹ - (1/3)x², we integrate term by term: 4x⁹ becomes 4x¹⁰/10 = (2/5)x¹⁰, and -(1/3)x² becomes -(1/3)x³/3 = -(1/9)x³. The antiderivative is (2/5)x¹⁰ - (1/9)x³ + C. Choice D ((2/5)x¹⁰ - (1/6)x³ + C) incorrectly computes the integral of -(1/3)x² as -(1/6)x³, perhaps by adding denominators instead of multiplying fractions. Remember: when integrating (a/b)xⁿ, the result is (a/b)·xⁿ⁺¹/(n+1) = a·xⁿ⁺¹/[b(n+1)].
The instantaneous power is $P(t)=9t^4-12t^3$ . Which is an antiderivative of $P(t)$ ?
Answer: \frac{9}{5}t^5- $3t^4$ +C$ This question tests the skill of finding antiderivatives by reversing the power rule for differentiation. To find the antiderivative of P(t) = 9t⁴ - 12t³, increase the exponent of each term by one and divide by the new exponent, resulting in (9/5)t⁵ - (12/4)t⁴ + C. This process reverses differentiation because differentiating (9/5)t⁵ - 3t⁴ + C yields 9t⁴ - 12t³, matching the original power function. The constant C accounts for the fact that differentiation eliminates constants, so antiderivatives differ by a constant. A tempting distractor like choice C fails because it incorrectly uses coefficients of 9 and -12 without dividing properly. Always remember that integration is the reverse of differentiation, so check your antiderivative by differentiating it back to see if you get the original function.
A particle's velocity is $v(t)=6t^2-4t+9$ . Which function is an antiderivative of $v(t)$ ?
Answer: $2t^3-2t^2+9t+C$ This problem requires finding an antiderivative 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. Combining these terms gives 2t³ - 2t² + 9t + C. Choice E (2t³ - 2t² + 9 + C) incorrectly integrates the constant 9 as 9 instead of 9t, forgetting that the integral of a constant is that constant times the variable. Remember: to verify an antiderivative, differentiate your answer and check if you get back the original function.
The rate of water flow is $r(t)=8t^3+5$ . Which function is an antiderivative of $r(t)$ ?
Answer: $2t^4+5t+C$ This problem tests your ability to find antiderivatives using basic integration techniques. To find an antiderivative of r(t) = 8t³ + 5, we integrate each term separately using the power rule: ∫tⁿ dt = $t^{n+1}$ /(n+1) + C. For 8t³, we get 8·t⁴/4 = 2t⁴, and for the constant 5, we get 5t (since ∫5 dt = 5t). Therefore, the antiderivative is 2t⁴ + 5t + C. Choice C (24t² + 5) represents the derivative of r(t), not its antiderivative, which reverses the student's thinking about the operation. When integrating, always increase the exponent by 1 and divide by the new exponent, and don't forget to add the constant of integration C.

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