Fundamental Theorem of Calculus: Definite Intervals - AP Calculus AB

Let $k(x)=\dfrac{2}{(x+1)^3}$ with antiderivative $K(x)=-(x+1)^{-2}$ ; compute $\int_{0}^{2} k(x)\,dx$ .
Answer: $K(2)-K(0)$ 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 2/ $(x+1)^3$ dx equals K(2) - K(0), where K is - $(x+1)^{-2}$ . K(2)= - $(3)^{-2}$ = -1/9, K(0)= - $(1)^{-2}$ =-1, so -1/9 - (-1)=8/9. This positive value makes sense. A tempting distractor is choice B, K(0) - K(2)=-1 - (-1/9)=-8/9, negating it. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.
Let $q(x)=\dfrac{3}{x}$ with antiderivative $Q(x)=3\ln x$ ; compute $\int_{1}^{e} q(x)\,dx$ .
Answer: $3\ln e-3\ln 1$ 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 e of 3/x dx equals Q(e) - Q(1), where Q is 3 ln x. Q(e)=3 ln e=3, Q(1)=3 ln1=0, so 3-0=3. This simplifies nicely due to log properties. A tempting distractor is choice A, 3 ln1 - 3 ln e =0-3=-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.
If $C'(x)=c(x)$ and $C(x)=\cos x-x$ , what is $\int_{\pi/2}^{\pi} c(x)\,dx$ ?
Answer: $C(\pi)-C\!\left(\frac\pi2\right)$ This question tests the Fundamental Theorem of Calculus Part 2, which states that if C'(x) = c(x), then ∫[a to b] c(x)dx = C(b) - C(a). With C(x) = cos x - x and the integral from π/2 to π, we need to evaluate C(π) - C(π/2). Computing: C(π) = cos(π) - π = -1 - π and C(π/2) = cos(π/2) - π/2 = 0 - π/2 = -π/2. Therefore, ∫[π/2 to π] c(x)dx = C(π) - C(π/2) = (-1 - π) - (-π/2) = -1 - π + π/2 = -1 - π/2. The distractor C(π/2) - C(π) would give -π/2 - (-1 - π) = -π/2 + 1 + π = 1 + π/2, incorrectly reversing the evaluation order. Always evaluate the antiderivative at the upper bound minus the lower bound to correctly apply FTC Part 2.
Given $F'(x)=f(x)$ and $F(x)=x^3-2x$ , what is $\int_{-1}^{2} f(x)\,dx$ ?
Answer: $F(2)-F(-1)$ 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 - 2x, we can directly evaluate the definite integral from -1 to 2. We calculate F(2) = 2³ - 2(2) = 8 - 4 = 4 and F(-1) = (-1)³ - 2(-1) = -1 + 2 = 1. Therefore, ∫[-1 to 2] f(x)dx = F(2) - F(-1) = 4 - 1 = 3. Students might incorrectly choose option B by reversing the subtraction order, forgetting that the FTC always uses upper bound minus lower bound. When applying FTC Part 2, always evaluate the antiderivative at the upper limit first, then subtract its value at the lower limit.
Given $n(x)=\dfrac{2}{1+x}$ and $N(x)=2\ln|1+x|$ , find $\int_{1}^{3} n(x)\,dx$ .
Answer: $N(3)-N(1)$ 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 2/(1+x) dx equals N(3) - N(1), where N is $2 \ln|1+x|$ . $N(3)=2 \ln 4$ , $N(1)=2 \ln 2$ , so $2 \ln 4 - 2 \ln 2=2 \ln(2)$ . This simplifies well. A tempting distractor is choice A, $N(1) - N(3)=2 \ln 2 - 2 \ln 4= -2 \ln 2$ , negating it. 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 $F(x)=x^3-4x$ , what is $\int_{-1}^{2} f(x)\,dx$ ?
Answer: $F(2)-F(-1)$ 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.
An antiderivative of $q$ is $Q(x)=x^4$ . What is $\int_{-2}^{1} q(x)\,dx$ ?
Answer: $Q(1)-Q(-2)$ This problem applies the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral over a negative-to-positive interval. Since Q(x) = x⁴ is an antiderivative of q(x), we have Q'(x) = q(x) = 4x³, so ∫[-2 to 1] q(x)dx = Q(1) - Q(-2). We calculate Q(1) = 1⁴ = 1 and Q(-2) = (-2)⁴ = 16. Therefore, ∫[-2 to 1] q(x)dx = Q(1) - Q(-2) = 1 - 16 = -15. Option C shows the reversed subtraction Q(-2) - Q(1) = 16 - 1 = 15, which has the wrong sign for this integral. The FTC Part 2 formula F(upper) - F(lower) applies even when the lower bound is negative, preserving the integral's true value.
If $f(x)=2x e^{x^2}$ and $F(x)=e^{x^2}$ , compute $\int_{0}^{2} f(x)\,dx$ by FTC.
Answer: $F(2)-F(0)$ This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, $\int_{0}^{2} 2x e^{x^2} \, dx$ equals $F(2) - F(0)$ , where $F$ is $e^{x^2}$ . $F(2)=e^4$ , $F(0)=e^0=1$ , so $e^4 - 1$ . This uses substitution implicitly. A tempting distractor is choice A, $F(0) - F(2)=1 - e^4$ , which is negative. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.
Given $w(x)=3x^2$ with antiderivative $W(x)=x^3$ , find $\int_{-2}^{1} w(x)\,dx$ .
Answer: $W(1)-W(-2)$ 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.
An antiderivative of $r(x)$ is $R(x)=x^4+5$ . Find $\int_{-1}^{1} r(x)\,dx$ using FTC.
Answer: $R(1)-R(-1)$ This problem demonstrates the Fundamental Theorem of Calculus Part 2, which states that the definite integral of a function equals the difference of its antiderivative evaluated at the bounds. Given that R(x) = x⁴ + 5 is an antiderivative of r(x), we evaluate ∫[-1 to 1] r(x)dx as R(1) - R(-1). Computing: R(1) = 1 + 5 = 6 and R(-1) = 1 + 5 = 6, so R(1) - R(-1) = 6 - 6 = 0. Option B, R(-1) - R(1), would also give 0 in this case, but it represents the incorrect order that would typically yield the wrong sign. The correct application of FTC Part 2 always evaluates as F(upper) - F(lower), maintaining proper orientation.
Let $f(x)=\cos x$ and an antiderivative be $F(x)=\sin x$ ; find $\int_{0}^{\pi/3} f(x)\,dx$ .
Answer: $\sin\left(\frac{\pi}{3}\right)-\sin 0$ 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.
Let $S'(x)=s(x)$ and $S(x)=x-\dfrac{1}{x}$ . Evaluate $\int_{1}^{3} s(x)\,dx$ .
Answer: $S(3)-S(1)$ This problem uses the Fundamental Theorem of Calculus Part 2 with a rational function antiderivative. Given that S'(x) = s(x) and S(x) = x - 1/x, we can find ∫[1 to 3] s(x)dx by computing S(3) - S(1). We calculate S(3) = 3 - 1/3 = 9/3 - 1/3 = 8/3 and S(1) = 1 - 1/1 = 1 - 1 = 0. Therefore, ∫[1 to 3] s(x)dx = S(3) - S(1) = 8/3 - 0 = 8/3. Choice D (s(3) - s(1)) incorrectly uses the derivative function rather than the antiderivative. The fundamental principle remains: evaluate the antiderivative at the upper bound and subtract the antiderivative at the lower bound.
Given $F'(x)=\dfrac{1}{x}$ and $F(x)=\ln x$ , evaluate $\int_{2}^{8} \dfrac{1}{x}\,dx$ using FTC Part 2.
Answer: $\ln 8-\ln 2$ 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 8 of (1/x) dx equals F(8) - F(2), where F is ln x. Evaluate F at the upper limit 8 to get ln 8, then subtract F at the lower limit 2, which is ln 2. This gives ln 8 - ln 2, equivalent to ln(8/2) by logarithm properties. A tempting distractor is choice A, ln 2 - ln 8, which switches the order and negates the value. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.
Let $f(x)=\dfrac{1}{(1-x)}$ with antiderivative $F(x)=-\ln|1-x|$ ; evaluate $\int_{0}^{1/2} f(x)\,dx$ .
Answer: $-\ln\left|1-\frac12\right|-\left(-\ln\left|1-0\right|\right)$ 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 1/2 of 1/(1-x) dx equals F(1/2) - F(0), where F is -ln|1-x|. F(1/2)=-ln|1-1/2|=-ln(1/2), F(0)=-ln|1-0|=-ln1=0, so -ln(1/2) - 0 = ln2. But choice B is -ln|1-1/2| - (-ln|1-0|), which is the same as F(1/2) - F(0). A tempting distractor is choice A, which swaps the order and gives -ln2. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.
If $S'(x)=s(x)$ and $S(x)=\tan x$ , find $\int_{0}^{\pi/4} s(x)\,dx$ .
Answer: $S(\pi/4)-S(0)$ This question requires using the Fundamental Theorem of Calculus Part 2 with a trigonometric antiderivative. Given that S'(x) = s(x) and S(x) = tan(x), we evaluate ∫[0 to π/4] s(x)dx = S(π/4) - S(0). We calculate S(π/4) = tan(π/4) = 1 and S(0) = tan(0) = 0. Therefore, ∫[0 to π/4] s(x)dx = S(π/4) - S(0) = 1 - 0 = 1. Option B incorrectly reverses the subtraction to S(0) - S(π/4) = 0 - 1 = -1, which would be negative when integrating the positive function sec²(x) over this interval. Always apply FTC Part 2 as F(upper limit) - F(lower limit) to get the correct sign and value.
A particle's velocity is $v(t)=3t^2-4t+1$ ; find $\int_{1}^{4} v(t)\,dt$ using an antiderivative $V(t)=t^3-2t^2+t$ .
Answer: $V(4)-V(1)$ 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.
Given $F'(x)=f(x)$ and $F(x)=3x^2-4x+1$ , find $\int_{-1}^{2} f(x)\,dx$ using FTC.
Answer: $F(2)-F(-1)$ 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) = 3x² - 4x + 1, we can directly evaluate the integral from -1 to 2. We calculate F(2) = 3(2)² - 4(2) + 1 = 12 - 8 + 1 = 5 and F(-1) = 3(-1)² - 4(-1) + 1 = 3 + 4 + 1 = 8. Therefore, ∫[-1 to 2] f(x)dx = F(2) - F(-1) = 5 - 8 = -3. Choice E (f(2) - f(-1)) is incorrect because it uses the derivative function instead of the antiderivative. Remember: FTC Part 2 always uses the antiderivative F evaluated at the upper bound minus the lower bound.
If $A'(x)=a(x)$ and $A(x)=\frac{1}{x}$ , what is $\int_{2}^{5} a(x)\,dx$ ?
Answer: $A(5)-A(2)$ This question applies the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral using a given antiderivative. Since A'(x) = a(x) and A(x) = 1/x, we can find ∫[2 to 5] a(x)dx = A(5) - A(2). We calculate A(5) = 1/5 and A(2) = 1/2. Therefore, ∫[2 to 5] a(x)dx = A(5) - A(2) = 1/5 - 1/2 = 2/10 - 5/10 = -3/10. Option A represents the common mistake of subtracting in reverse order, giving A(2) - A(5) = 1/2 - 1/5 = 3/10, which has the wrong sign. The FTC Part 2 formula F(b) - F(a) preserves the direction of integration, so always evaluate at the upper limit first.
If $P'(x)=p(x)$ and $P(x)=e^{2x}$ , evaluate $\int_{1}^{4} p(x)\,dx$ .
Answer: $P(4)-P(1)$ 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.
If $s(x)=2\sin x$ and $S(x)=-2\cos x$ is an antiderivative, find $\int_{\pi/2}^{\pi} s(x)\,dx$ .
Answer: $S(\pi)-S\left(\frac{\pi}{2}\right)$ 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 π/2 to π of 2 sin x dx equals S(π) - S(π/2), where S is -2 cos x. S(π)=-2 cosπ= -2(-1)=2, S(π/2)=-2 cos(π/2)=0, so 2-0=2. This matches the integral value. A tempting distractor is choice A, S(π/2) - S(π)=0-2=-2, which negates it. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.

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