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AP Calculus AB Quiz
Practice Fundamental Theorem Of Calculus Definite Intervals in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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Let f(x)=x and F(x)=32x3/2; evaluate ∫04f(x)dx.
This quiz focuses on Fundamental Theorem Of Calculus Definite Intervals, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.
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.
Let f(x)=x and F(x)=32x3/2; evaluate ∫04f(x)dx.
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from 0 to 4 of √x dx equals F(4) - F(0), where F is (2/3)x^{3/2}. F(4)=(2/3)(8)=16/3, F(0)=0, so 16/3 - 0=16/3. This gives the area. A tempting distractor is choice A, F(0) - F(4)=0 - 16/3=-16/3, reversing the sign. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.
Given w(x)=3x2 with antiderivative W(x)=x3, find ∫−21w(x)dx.
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from -2 to 1 of 3x^2 dx equals W(1) - W(-2), where W is x^3. W(1)=1, W(-2)=-8, so 1 - (-8)=9. Note the limits: upper 1, lower -2. A tempting distractor is choice B, W(-2) - W(1)=-8-1=-9, reversing the order. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.
A particle’s velocity is v(t)=3t2−4t+1; find ∫14v(t)dt using an antiderivative V(t)=t3−2t2+t.
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from 1 to 4 of v(t) dt equals V(4) - V(1), where V is the given antiderivative. To compute this, evaluate V at the upper limit 4 and subtract V at the lower limit 1. This direct substitution yields the net change in the antiderivative over the interval. A tempting distractor is choice B, V(1) - V(4), which reverses the order and results in the negative of the correct integral. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.
A force is F(x)=4x; with antiderivative A(x)=38x3/2, find ∫19F(x)dx.
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from 1 to 9 of 4xdx equals A(9)−A(1), where A(x)=38x3/2. Evaluate A(9)=38×27=72, A(1)=38×1=38, so 72−38=3216−8=3208. This gives the work done by the force. A tempting distractor is choice A, A(1)−A(9), which negates the correct value. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.
Let f(x)=cosx and an antiderivative be F(x)=sinx; find ∫0π/3f(x)dx.
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from 0 to π/3 of cos x dx equals F(π/3) - F(0), where F is sin x. Compute sin(π/3) which is √3/2, and subtract sin(0) which is 0. This results in √3/2 - 0 = √3/2, the area under the curve. A tempting distractor is choice A, sin 0 - sin(π/3), which reverses subtraction and gives a negative result. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.
If g(x)=6x−5 and G(x)=3x2−5x is an antiderivative, find ∫−13g(x)dx.
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from -1 to 3 of (6x-5) dx equals G(3) - G(-1), where G is 3x^2 - 5x. Compute G(3) = 3(9) - 5(3) = 12, and G(-1) = 3(1) - 5(-1) = 8, so 12 - 8 = 4. This calculation gives the correct integral value. A tempting distractor is choice A, G(-1) - G(3), which reverses the order and yields -4. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.
Given h(x)=sec2x with antiderivative H(x)=tanx, evaluate ∫0π/6h(x)dx.
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from 0 to π/6 of sec^2 x dx equals H(π/6) - H(0), where H is tan x. Evaluate tan(π/6) = 1/√3, and tan(0) = 0, so 1/√3 - 0. This provides the exact value of the integral. A tempting distractor is choice B, tan 0 - tan(π/6), which switches subtraction and gives a negative. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.
If F′(x)=f(x) and F(x)=x3−4x, what is ∫−12f(x)dx?
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2, which states that if F'(x) = f(x), then ∫[a to b] f(x)dx = F(b) - F(a). Since we're given that F'(x) = f(x) and F(x) = x³ - 4x, we can directly apply the theorem. To evaluate ∫[-1 to 2] f(x)dx, we compute F(2) - F(-1). We find F(2) = 2³ - 4(2) = 8 - 8 = 0 and F(-1) = (-1)³ - 4(-1) = -1 + 4 = 3, giving us F(2) - F(-1) = 0 - 3 = -3. A common error would be reversing the order to get F(-1) - F(2) = 3, which incorrectly switches the bounds. Remember: when using FTC Part 2, always evaluate the antiderivative at the upper bound minus the antiderivative at the lower bound.
A tank’s inflow rate is r(t)=2et; with antiderivative R(t)=2et, compute ∫02r(t)dt.
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from 0 to 2 of 2e^t dt equals R(2) - R(0), where R is 2e^t. Evaluate R at 2 to get 2e^2, then subtract R at 0 which is 2e^0 = 2. This yields 2e^2 - 2, the net accumulation. A tempting distractor is choice A, 2e^0 - 2e^2, which inverts the limits and negates the integral. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.
If H′(x)=h(x) and H(x)=sinx+2x, what is ∫0πh(x)dx?
Explanation: This question applies the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral using a known antiderivative. Given that H'(x) = h(x) and H(x) = sin x + 2x, we can find ∫[0 to π] h(x)dx = H(π) - H(0). Evaluating at the bounds: H(π) = sin(π) + 2π = 0 + 2π = 2π and H(0) = sin(0) + 2(0) = 0 + 0 = 0. Thus, ∫[0 to π] h(x)dx = H(π) - H(0) = 2π - 0 = 2π. The distractor H(0) - H(π) = -2π incorrectly reverses the evaluation order, yielding a negative result when the actual integral is positive. To correctly apply FTC Part 2, always evaluate F(upper bound) - F(lower bound), maintaining the proper order to get the correct sign.
An antiderivative of p(x) is P(x)=x1−x2. Evaluate ∫13p(x)dx using FTC.
Explanation: This problem demonstrates the Fundamental Theorem of Calculus Part 2, which states that for an antiderivative P(x) of p(x), the definite integral ∫[a to b] p(x)dx equals P(b) - P(a). Given P(x) = 1/x - x², we need to evaluate ∫[1 to 3] p(x)dx = P(3) - P(1). Computing the values: P(3) = 1/3 - 3² = 1/3 - 9 = 1/3 - 27/3 = -26/3 and P(1) = 1/1 - 1² = 1 - 1 = 0. Therefore, ∫[1 to 3] p(x)dx = P(3) - P(1) = -26/3 - 0 = -26/3. The incorrect choice P(1) - P(3) = 0 - (-26/3) = 26/3 reverses the bounds and changes the sign of the result. When using FTC Part 2, maintain the order: antiderivative at upper bound minus antiderivative at lower bound.
Given T′(x)=t(x) and T(x)=2x2−5, what is ∫−33t(x)dx?
Explanation: This problem demonstrates the Fundamental Theorem of Calculus Part 2 applied to a symmetric interval. Since T'(x) = t(x) and T(x) = x²/2 - 5, we can find ∫[-3 to 3] t(x)dx = T(3) - T(-3). We calculate T(3) = 3²/2 - 5 = 9/2 - 5 = 4.5 - 5 = -0.5 and T(-3) = (-3)²/2 - 5 = 9/2 - 5 = 4.5 - 5 = -0.5. Therefore, ∫[-3 to 3] t(x)dx = T(3) - T(-3) = -0.5 - (-0.5) = 0. Option D represents the incorrect order T(-3) - T(3), but in this case would also equal 0 since T(3) = T(-3). The key insight is recognizing that t(x) = T'(x) = x is an odd function, so its integral over a symmetric interval is zero.
If A′(x)=a(x) and A(x)=ex−5, what is ∫−20a(x)dx?
Explanation: This question requires applying the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral using the given antiderivative relationship. Since A'(x) = a(x) and A(x) = eˣ - 5, we can find ∫[-2 to 0] a(x)dx = A(0) - A(-2). Evaluating the antiderivative at the bounds: A(0) = e⁰ - 5 = 1 - 5 = -4 and A(-2) = e⁻² - 5 = 1/e² - 5. Therefore, ∫[-2 to 0] a(x)dx = A(0) - A(-2) = -4 - (1/e² - 5) = -4 - 1/e² + 5 = 1 - 1/e². The distractor A(-2) - A(0) would incorrectly reverse the evaluation, giving 1/e² - 1 instead. Remember to always evaluate the antiderivative at the upper limit minus the antiderivative at the lower limit when applying FTC Part 2.
If N′(x)=n(x) and N(x)=x4+1, evaluate ∫2−1n(x)dx using FTC.
Explanation: This question requires careful application of the Fundamental Theorem of Calculus Part 2 with reversed bounds of integration. Given N'(x) = n(x) and N(x) = x⁴ + 1, we need to evaluate ∫[2 to -1] n(x)dx = N(-1) - N(2). Note that the upper bound is -1 and the lower bound is 2, so we compute N(-1) - N(2). Evaluating: N(-1) = (-1)⁴ + 1 = 1 + 1 = 2 and N(2) = 2⁴ + 1 = 16 + 1 = 17. Therefore, ∫[2 to -1] n(x)dx = N(-1) - N(2) = 2 - 17 = -15. The distractor N(2) - N(-1) = 17 - 2 = 15 incorrectly treats 2 as the upper bound, yielding the wrong sign. When bounds appear reversed (upper < lower), still apply FTC Part 2 as F(upper) - F(lower), which naturally produces the correct negative result.
An antiderivative of m(x) is M(x)=tan−1(x). What is ∫01m(x)dx?
Explanation: This problem demonstrates the Fundamental Theorem of Calculus Part 2, which relates definite integrals to antiderivatives through the evaluation formula. Given that M(x) = tan⁻¹(x) is an antiderivative of m(x), we can find ∫[0 to 1] m(x)dx = M(1) - M(0). Evaluating: M(1) = tan⁻¹(1) = π/4 (since tan(π/4) = 1) and M(0) = tan⁻¹(0) = 0 (since tan(0) = 0). Therefore, ∫[0 to 1] m(x)dx = M(1) - M(0) = π/4 - 0 = π/4. The incorrect choice M(0) - M(1) = 0 - π/4 = -π/4 reverses the bounds and produces a negative value when the integral should be positive. To correctly apply FTC Part 2, always compute F(upper) - F(lower) to maintain the proper sign.
Given B(x)=x+21x is an antiderivative of b(x), find ∫49b(x)dx.
Explanation: This problem applies the Fundamental Theorem of Calculus Part 2, which connects antiderivatives to definite integrals through the formula ∫[a to b] f(x)dx = F(b) - F(a). Given that B(x) = √x + (1/2)x is an antiderivative of b(x), we evaluate ∫[4 to 9] b(x)dx = B(9) - B(4). Computing the values: B(9) = √9 + (1/2)(9) = 3 + 4.5 = 7.5 and B(4) = √4 + (1/2)(4) = 2 + 2 = 4. Therefore, ∫[4 to 9] b(x)dx = B(9) - B(4) = 7.5 - 4 = 3.5. The incorrect choice B(4) - B(9) = 4 - 7.5 = -3.5 reverses the proper order and yields a negative result for what should be positive. When applying FTC Part 2, always subtract the lower bound evaluation from the upper bound evaluation.
If S′(x)=s(x) and S(x)=x+x, evaluate ∫49s(x)dx.
Explanation: This question applies the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral using a known antiderivative relationship. Since S'(x) = s(x) and S(x) = √x + x, we compute ∫[4 to 9] s(x)dx as S(9) - S(4). Evaluating at the bounds: S(9) = √9 + 9 = 3 + 9 = 12 and S(4) = √4 + 4 = 2 + 4 = 6, so S(9) - S(4) = 12 - 6 = 6. Option A, S(4) - S(9), would give -6, which has the wrong sign due to incorrect ordering of the subtraction. Remember that FTC Part 2 always uses the pattern: antiderivative at upper bound minus antiderivative at lower bound.
A particle’s velocity is v(t) with antiderivative V(t)=t3−6t. Find ∫−21v(t)dt.
Explanation: This problem uses the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral of velocity. Since V(t) is an antiderivative of v(t) (meaning V'(t) = v(t)), and V(t) = t³ - 6t, we can find ∫[-2 to 1] v(t)dt by calculating V(1) - V(-2). We compute V(1) = 1³ - 6(1) = 1 - 6 = -5 and V(-2) = (-2)³ - 6(-2) = -8 + 12 = 4. Therefore, ∫[-2 to 1] v(t)dt = V(1) - V(-2) = -5 - 4 = -9. Choice A (V(-2) - V(1)) reverses the order, which would give the negative of the correct answer. When applying FTC Part 2, always subtract the antiderivative at the lower bound from the antiderivative at the upper bound.
Given T′(x)=t(x) and T(x)=arctanx, compute ∫−11t(x)dx using FTC.
Explanation: This problem applies the Fundamental Theorem of Calculus Part 2 using an inverse trigonometric antiderivative. Since T'(x) = t(x) and T(x) = arctan x, we can find ∫[-1 to 1] t(x)dx by computing T(1) - T(-1). We calculate T(1) = arctan(1) = π/4 and T(-1) = arctan(-1) = -π/4. Therefore, ∫[-1 to 1] t(x)dx = T(1) - T(-1) = π/4 - (-π/4) = π/4 + π/4 = π/2. Choice C (t(1) - t(-1)) incorrectly uses the derivative function instead of the antiderivative. Remember that FTC Part 2 always requires evaluating the antiderivative function at the bounds, not the integrand itself.
If P′(x)=p(x) and P(x)=e2x, evaluate ∫14p(x)dx.
Explanation: This problem applies the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral using an exponential antiderivative. Since P'(x) = p(x) and P(x) = e^(2x), we can find ∫[1 to 4] p(x)dx by calculating P(4) - P(1). We compute P(4) = e^(2·4) = e^8 and P(1) = e^(2·1) = e^2. Therefore, ∫[1 to 4] p(x)dx = P(4) - P(1) = e^8 - e^2. Choice C (p(4) - p(1)) incorrectly uses the derivative function instead of the antiderivative, a common misconception. Remember that FTC Part 2 always involves the antiderivative function evaluated at the bounds, not the original function being integrated.