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Finding Antiderivatives and Indefinite Integrals - AP Calculus BC

  1. A particle's velocity is v(t)=6t2−4t+5v(t)=6t^2-4t+5. Which function is an antiderivative of v(t)v(t)?
    Answer: 2t3−2t2+5t+C2t^3-2t^2+5t+C This problem requires finding an antiderivative of the velocity function v(t) = 6t² - 4t + 5. To find an antiderivative, we reverse the differentiation process by applying the power rule in reverse: increase each exponent by 1 and divide by the new exponent. 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 C shows the derivative of v(t) rather than its antiderivative, a common error. Remember: when finding antiderivatives, always add 1 to the exponent and divide by the new exponent, then include +C.
  2. The rate of water flow is r(x)=5x−2−7x3r(x)=5x^{-2}-7x^{3}. Which function is an antiderivative of r(x)r(x)?
    Answer: −5x−74x4+C-\frac{5}{x}-\frac{7}{4}x^4+C This problem requires finding antiderivatives using the power rule for negative exponents. To integrate r(x) = 5x⁻² - 7x³, we apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C for n ≠ -1. For 5x⁻², we get 5x⁻¹/(-1) = -5x⁻¹ = -5/x; for -7x³, we get -7x⁴/4 = -7x⁴/4. The complete antiderivative is -5/x - 7x⁴/4 + C. Choice C (-5x⁻¹ - 7x⁴ + C) might seem correct but contains an error: it shows -7x⁴ instead of -7x⁴/4, forgetting to divide by the new exponent. When integrating powers of x, always remember to add 1 to the exponent and divide by the new exponent, being especially careful with negative exponents.
  3. A function has derivative g′(x)=−5x−2+11g'(x)=-5x^{-2}+11. Which is an antiderivative of g′(x)g'(x)?
    Answer: 5x−1+11x+C5x^{-1}+11x+C This problem tests antiderivative skills with negative exponents and constants. To find an antiderivative of g'(x) = -5x⁻² + 11, we integrate term by term: for -5x⁻², we increase the exponent to -1 and divide: -5x⁻¹/(-1) = 5x⁻¹. The constant 11 integrates to 11x. Adding the constant of integration gives us 5x⁻¹ + 11x + C. Choice D shows -5x⁻¹ + 11x + C, which has the wrong sign on the first term—this occurs when students forget that dividing by -1 changes the sign. When integrating negative powers, remember that xⁿ⁺¹/(n+1) can change signs: when n = -2, you divide by -1, which flips the sign of the coefficient.
  4. A cost gradient is g(x)=3x+4x−3g(x)=\dfrac{3}{x}+4x^{-3}. Which expression is an antiderivative of g(x)g(x)?
    Answer: 3ln⁡∣x∣−2x−2+C3\ln|x|-2x^{-2}+C This problem requires finding an antiderivative of g(x) = 3/x + 4x⁻³. To find antiderivatives, we reverse differentiation for each term. For 3/x, the antiderivative is 3ln|x|. For 4x⁻³, we add 1 to get x⁻², and dividing by -2 gives us -2x⁻². Including the constant of integration yields 3ln|x| - 2x⁻² + C. Choice A incorrectly shows +2x⁻² instead of -2x⁻², missing the sign from integrating a negative power. Remember: when integrating x⁻ⁿ where n > 1, the coefficient changes sign because you divide by a negative number.
  5. A current is modeled by I(t)=8t−3+5t2I(t)=8t^{-3}+5t^{2}. Which function is an antiderivative of I(t)I(t)?
    Answer: −4t−2+53t3+C-4t^{-2}+\dfrac{5}{3}t^{3}+C This problem tests integration with negative exponents, requiring careful sign tracking. To find an antiderivative of I(t) = 8t⁻³ + 5t², we integrate each term: for 8t⁻³, we get 8t⁻²/(-2) = -4t⁻². For 5t², we get 5t³/3 = (5/3)t³. The complete antiderivative is -4t⁻² + (5/3)t³ + C. Choice D shows 4t⁻² + (5/3)t³ + C with the wrong sign on the first term—this error occurs when students forget that integrating t⁻³ involves dividing by -2, which introduces a negative sign. When integrating negative powers of t, always check whether n+1 is positive or negative: if n < -1, then n+1 is still negative, creating a sign change in the result.
  6. A spring's force is modeled by F(x)=4−10xF(x)=4-10x. Which function is an antiderivative of F(x)F(x)?
    Answer: 4x−5x2+C4x-5x^{2}+C 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.
  7. The acceleration is a(t)=7et−4t2a(t)=7e^t-\dfrac{4}{t^2}. Which function is an antiderivative of a(t)a(t)?
    Answer: 7et+4t+C7e^t+\dfrac{4}{t}+C 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.
  8. A population model has rate P′(t)=4e−t+t−2P'(t)=4\text{e}^{-t}+t^{-2}. Which could be P(t)P(t)?
    Answer: −4e−t−1t+C-4\text{e}^{-t}-\frac{1}{t}+C This problem requires finding an antiderivative involving exponential and power functions. Given P'(t) = 4e⁻ᵗ + t⁻², we integrate each term separately. For 4e⁻ᵗ, using the rule ∫eᵃᵗ dt = (1/a)eᵃᵗ + C with a = -1, we get 4·(1/-1)e⁻ᵗ = -4e⁻ᵗ. For t⁻², applying the power rule gives t⁻¹/(-1) = -t⁻¹ = -1/t. The antiderivative is -4e⁻ᵗ - 1/t + C. Choice B might mislead students who forget that integrating e⁻ᵗ introduces a negative sign due to the chain rule in reverse. When integrating exponential functions with negative exponents, remember to divide by the coefficient in the exponent, including its sign.
  9. A chemical concentration changes at rate c′(t)=2t−3+7c'(t)=2t^{-3}+7. Which expression is an antiderivative of c′(t)c'(t)?
    Answer: −t−2+7t+C-t^{-2}+7t+C This problem tests basic antiderivative reasoning skills. For c'(t) = 2t−32t^{-3} + 7, we reverse differentiation using the power rule. Rewrite as 2t−32t^{-3} + 7t07t^0, then add one to -3 for t−2t^{-2} and divide by -2, giving -t−2t^{-2} after multiplying by 2/-2. For 7, it integrates to 7t. Choice B distracts by using -2t−22t^{-2}, forgetting to simplify the coefficient properly. Always simplify fractions after applying the power rule and include +C for indefinite integrals.
  10. A temperature model has T′(t)=7t6+ 2t3T'(t)=7t^6+\,\frac{2}{t^3}. Which expression is an antiderivative of T′(t)T'(t)?​
    Answer: t7−1t2+Ct^7-\frac{1}{t^2}+C 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.
  11. A tank fills at rate r(t)=8−2t−3r(t)=8-2t^{-3} liters per minute; which function is an antiderivative of r(t)r(t)?
    Answer: 8t+t−2+C8t+t^{-2}+C Finding antiderivatives involves applying basic integration rules to reverse the process of differentiation. To find the antiderivative of (r(t) = 8 - 2t−32t^{-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−3-2t^{-3}) is (t−2t^{-2}), resulting in (8t + t−2t^{-2} + C). This reverses differentiation because differentiating (8t + t−2t^{-2} + C) gives back (8 - 2t−32t^{-3}). A tempting distractor like choice B fails because it incorrectly assigns a negative sign to (t−2t^{-2}), mishandling the integration of the negative coefficient. Always remember to add the constant of integration (C) and verify by differentiating the result.
  12. The derivative of a function is f′(x)=9e2x+2xf'(x)=9\text{e}^{2x}+2x. Which could be f(x)f(x)?
    Answer: 92e2x+x2+C\frac{9}{2}\text{e}^{2x}+x^{2}+C This problem involves finding an antiderivative that includes an exponential function with a coefficient in the exponent. Given f'(x) = 9e²ˣ + 2x, we need to integrate each term. For 2x, the antiderivative is x². For 9e²ˣ, we use the rule ∫eᵃˣ dx = (1/a)eᵃˣ + C, so ∫9e²ˣ dx = 9·(1/2)e²ˣ = (9/2)e²ˣ. The complete antiderivative is (9/2)e²ˣ + x² + C. Choice B might tempt students who forget to divide by the coefficient in the exponent when integrating exponential functions. Remember: when integrating eᵃˣ, always divide by the coefficient a in the exponent.
  13. A particle's velocity is v(t)=6t2−4t+9v(t)=6t^2-4t+9. Which expression is an antiderivative of v(t)v(t)?
    Answer: 2t3−2t2+9t+C2t^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, where C is the constant of integration. Choice B might seem tempting as it omits the constant C, but all antiderivatives must include this arbitrary constant since differentiation eliminates constants. When finding antiderivatives, always increase the power by 1 and divide by the new power, then add the constant of integration.
  14. A force function is F(x)=5x4+11x−9F(x)=5x^4+11x-9. Which function is an antiderivative of F(x)F(x)?​
    Answer: x5+112x2−9x+Cx^5+\frac{11}{2}x^2-9x+C 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.
  15. The slope of a curve is dydx=2x−6x\frac{dy}{dx}=2x-\frac{6}{x}. Which gives a family of antiderivatives?​
    Answer: x2−6ln⁡∣x∣+Cx^2-6\ln|x|+C Finding an antiderivative of dy/dx = 2x - 6/x gives the family of curves y(x). We integrate each term: ∫2x dx = 2x²/2 = x², and ∫(-6/x) dx = -6ln|x|. Adding the constant of integration yields x² - 6ln|x| + C, which represents all possible curves with the given slope. The term 1/x has the special antiderivative ln|x|, not a power function. Choice D shows -6/x, which would be the antiderivative of 6/x² rather than 6/x, a common confusion between these reciprocal forms. Remember that ∫(1/x) dx = ln|x| + C is an exception to the power rule and must be memorized separately.
  16. A revenue rate is R′(x)= 12x+5x4R'(x)=\,\frac{12}{\sqrt{x}}+5x^4. Which is an antiderivative of R′(x)R'(x)?​
    Answer: 24x+x5+C24\sqrt{x}+x^5+C This problem requires integrating a function with both a negative fractional exponent and a polynomial term. Rewriting R'(x) = 12x⁻¹/² + 5x⁴, we integrate: ∫12x⁻¹/² dx = 12x¹/²/(1/2) = 24x¹/² = 24√x, and ∫5x⁴ dx = 5x⁵/5 = x⁵. The complete antiderivative is 24√x + x⁵ + C. The key insight is recognizing that x⁻¹/² = 1/√x integrates to 2√x, then multiplied by 12 gives 24√x. Choice D incorrectly shows 5x⁵, keeping the coefficient 5 instead of dividing by the new exponent. Remember that when integrating xⁿ, the coefficient must be divided by (n+1), regardless of whether n is positive, negative, or fractional.
  17. The marginal cost is C′(x)=9x2+8x−1/2C'(x)=9x^2+8x^{-1/2}. Which expression is an antiderivative of C′(x)C'(x)?​
    Answer: 3x3+16x1/2+C3x^3+16x^{1/2}+C 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.
  18. The rate of water flow is r(t)=5t4−3t−2r(t)=5t^{4}-3t^{-2}. Find an antiderivative of r(t)r(t) with respect to tt.
    Answer: t5+3t−1+Ct^{5}+3t^{-1}+C This problem tests basic antiderivative reasoning skills. The antiderivative of r(t) = 5t⁴ - 3t−23t^{-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−23t^{-2}, increase to -1 and divide by -1, resulting in 3t−13t^{-1}. Choice B might tempt if one forgets to change the sign when integrating the negative term, but it incorrectly has -3t−13t^{-1}. A key strategy for integrating power functions is to handle negative exponents carefully and always add +C.
  19. An object's acceleration is a(t)=8−3t2a(t)=8-3t^{2}. Which function is an antiderivative of a(t)a(t) with respect to tt?
    Answer: 8t−t3+C8t-t^{3}+C This problem tests basic antiderivative reasoning skills. To find an antiderivative of a(t) = 8 - 3t², we reverse differentiation by integrating each term separately. The constant 8 integrates to 8t, as the antiderivative of a constant is the constant times t. For -3t², add one to the exponent to get t³ and divide by 3, yielding -t³. Choice A fails as a distractor because it keeps the coefficient -3 for t³ instead of dividing properly. For transferable success, integrate constants and powers term by term, ensuring to simplify coefficients correctly and add +C.
  20. A particle's velocity is v(t)=6t2−4t+5v(t)=6t^2-4t+5. Which function is an antiderivative of v(t)v(t)?​
    Answer: 2t3−2t2+5t+C2t^3-2t^2+5t+C 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.

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