Fundamental Theorem of Calculus: Definite Intervals
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AP Calculus AB › Fundamental Theorem of Calculus: Definite Intervals
Let $c(x)=\dfrac{1}{(x+2)}$ with antiderivative $C(x)=\ln|x+2|$; evaluate $\int_{0}^{3} c(x),dx$.
$\ln|5|$
$\ln|2|$
$\ln|2|-\ln|5|$
$\ln|5|-\ln|2|$
$\ln|5|+\ln|2|$
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 3 of 1/(x+2) dx equals C(3) - C(0), but C(x)=ln|x+2|, so C(3)=ln5, C(0)=ln2, ln5 - ln2=ln(5/2). Choices use ln|5| - ln|2|. A tempting distractor is choice A, ln|2| - ln|5|=ln(2/5), negative. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.
If $r(x)=\dfrac{1}{\sqrt{x}}$ and $R(x)=2\sqrt{x}$, evaluate $\int_{4}^{9} r(x),dx$.
$2\sqrt{4}$
$2\sqrt{9}$
$2\sqrt{9}-2\sqrt{4}$
$2\sqrt{4}-2\sqrt{9}$
$2\sqrt{9}+2\sqrt{4}$
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 4 to 9 of 1/√x dx equals R(9) - R(4), where R is 2√x. R(9)=23=6, R(4)=22=4, so 6-4=2. This is the area under the curve. A tempting distractor is choice A, 2√4 - 2√9=4-6=-2, which reverses subtraction. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.
If $a(x)=\sin x+\cos x$ and $A(x)=-\cos x+\sin x$, compute $\int_{0}^{\pi/2} a(x),dx$.
$A\left(\frac{\pi}{2}\right)$
$A\left(\frac{\pi}{2}\right)-A(0)$
$A\left(\frac{\pi}{2}\right)+A(0)$
$A(0)-A\left(\frac{\pi}{2}\right)$
$A(0)$
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/2 of (sin x + cos x) dx equals A(π/2) - A(0), where A is -cos x + sin x. A(π/2)=-cos(π/2) + sin(π/2)=0+1=1, A(0)=-cos0 + sin0=-1+0=-1, so 1 - (-1)=2. This sums the integrals. A tempting distractor is choice A, A(0) - A(π/2)=-1 -1=-2, reversing it. 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 $v(t)$ with antiderivative $V(t)=t^3-6t$. Find $\int_{-2}^{1} v(t),dt$.
$V(1)+V(-2)$
$V(2)-V(1)$
$V(1)-V(-2)$
$V(-2)-V(1)$
$v(1)-v(-2)$
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.
Let $S'(x)=s(x)$ and $S(x)=x-\dfrac{1}{x}$. Evaluate $\int_{1}^{3} s(x),dx$.
$s(3)-s(1)$
$S(3)+S(1)$
$S(3)-S(0)$
$S(3)-S(1)$
$S(1)-S(3)$
Explanation
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.
A force is $F(x)=4\sqrt{x}$; with antiderivative $A(x)=\dfrac{8}{3}x^{3/2}$, find $\int_{1}^{9} F(x),dx$.
$A(9)$
$A(9)+A(1)$
$A(1)-A(9)$
$A(9)-A(1)$
$A(1)$
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 $4\sqrt{x} , dx$ equals $A(9) - A(1)$, where $A(x) = \dfrac{8}{3} x^{3/2}$. Evaluate $A(9) = \dfrac{8}{3} \times 27 = 72$, $A(1) = \dfrac{8}{3} \times 1 = \dfrac{8}{3}$, so $72 - \dfrac{8}{3} = \dfrac{216 - 8}{3} = \dfrac{208}{3}$. 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.
Given $f(x)=5$ and antiderivative $F(x)=5x$, evaluate $\int_{-2}^{4} f(x),dx$ using FTC.
$F(-2)-F(4)$
$F(-2)$
$F(4)-F(-2)$
$F(4)$
$F(4)+F(-2)$
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 4 of 5 dx equals F(4) - F(-2), where F is 5x. F(4)=20, F(-2)=-10, so 20 - (-10)=30. This constant integral is straightforward. A tempting distractor is choice A, F(-2) - F(4)=-10-20=-30, 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.
Let $f(x)=\cos x$ and an antiderivative be $F(x)=\sin x$; find $\int_{0}^{\pi/3} f(x),dx$.
$\sin 0-\sin\left(\frac{\pi}{3}\right)$
$\sin\left(\frac{\pi}{3}\right)$
$\sin 0$
$\sin\left(\frac{\pi}{3}\right)+\sin 0$
$\sin\left(\frac{\pi}{3}\right)-\sin 0$
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 $s(x)=2\sin x$ and $S(x)=-2\cos x$ is an antiderivative, find $\int_{\pi/2}^{\pi} s(x),dx$.
$S(\pi)$
$S(\pi)+S\left(\frac{\pi}{2}\right)$
$S\left(\frac{\pi}{2}\right)-S(\pi)$
$S(\pi)-S\left(\frac{\pi}{2}\right)$
$S\left(\frac{\pi}{2}\right)$
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 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.
Let $f(x)=\sqrt{x}$ and $F(x)=\dfrac{2}{3}x^{3/2}$; evaluate $\int_{0}^{4} f(x),dx$.
$F(0)-F(4)$
$F(0)$
$F(4)-F(0)$
$F(4)+F(0)$
$F(4)$
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.