Fundamental Theorem of Calculus: Accumulation Functions - AP Calculus AB

1. Suppose $U(x)=\int_{1}^{x}\dfrac{e^{t}}{t}\,dt$ for $x>0$. What is $U'(x)$?
   Answer: $\dfrac{e^{x}}{x}$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function U(x) is defined as the integral from 1 to x of eᵗ / t dt, which accumulates the area under the curve of f(t) = eᵗ / t from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative U'(x) equals the integrand evaluated at x, which is eˣ / x. This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the integral of eᵗ / t dt evaluated from 1 to x, but that actually computes U(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.
2. Let $N(x)=\int_{1}^{x}\dfrac{1}{(1+t^2)^2}\,dt$. What is $N'(x)$?
   Answer: $\dfrac{1}{(1+x^2)^2}$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function N(x) is defined as the integral from 1 to x of 1/$(1+t^2)^2$ dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative N'(x) is simply the integrand evaluated at t = x, so N'(x) = 1/$(1+x^2)^2$. This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.
3. If $T(x)=\int_{-3}^{x}\left(e^{2t}-\cos t\right)dt$, what is $T'(x)$?
   Answer: $e^{2x}-\cos x$ This problem requires applying the Fundamental Theorem of Calculus Part 1, where the derivative of an integral with variable upper limit equals the integrand evaluated at that limit. Given $T(x) = \int_{-3}^{x}(e^{2t} - \cos t)\,dt$, we get $T'(x) = e^{2x} - \cos x$ by replacing t with x in the integrand. The process is purely evaluation; no integration or differentiation of the integrand is performed. Choice A ($2e^{2x} + \sin x$) shows the derivatives of $e^{2x}$ and $-\cos x$, which would result from incorrectly differentiating after applying FTC Part 1. To master this concept, remember that FTC Part 1 converts "derivative of integral" problems into simple substitution exercises.
4. Suppose $F(x)=\int_{0}^{x}\sqrt{9-t^2}\,dt$. What is $F'(x)$?
   Answer: $\sqrt{9-x^2}$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function F(x) is defined as the integral from 0 to x of √(9 - t²) dt, which accumulates the area under the curve of f(t) = √(9 - t²) from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative F'(x) equals the integrand evaluated at x, which is √(9 - x²). This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the evaluation of an antiderivative from 0 to x, but that actually computes F(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.
5. If $S(x)=\int_{2}^{x}\dfrac{t}{\sqrt{t^2+9}}\,dt$, what is $S'(x)$?
   Answer: $\dfrac{x}{\sqrt{x^2+9}}$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function S(x) is defined as the integral from 2 to x of t / √(t² + 9) dt, which accumulates the area under the curve of f(t) = t / √(t² + 9) from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative S'(x) equals the integrand evaluated at x, which is x / √(x² + 9). This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the evaluation of an antiderivative from 2 to x, but that actually computes S(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.
6. Define $Q(x)=\int_{10}^{x}\dfrac{\ln t}{t}\,dt$ for $x>0$. What is $Q'(x)$?
   Answer: $\dfrac{\ln x}{x}$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function Q(x) is defined as the integral from 10 to x of (ln t)/t dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative Q'(x) is simply the integrand evaluated at t = x, so Q'(x) = (ln x)/x. This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.
7. Let $P(x)=\int_{-2}^{x}\dfrac{1}{\sqrt{t+3}}\,dt$ for $x>-3$. What is $P'(x)$?
   Answer: $\dfrac{1}{\sqrt{x+3}}$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function P(x) is defined as the integral from -2 to x of 1/sqrt(t+3) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative P'(x) is simply the integrand evaluated at t = x, so P'(x) = 1/sqrt(x+3). This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.
8. If $H(x)=\int_{0}^{x}(\sin t+t^2)\,dt$, what is $H'(x)$?​​
   Answer: $\sin x+x^2$ This problem applies the Fundamental Theorem of Calculus Part 1, which tells us that differentiating an integral with variable upper limit gives the integrand evaluated at that limit. For $H(x) = \int_{0}^{x}(\sin t + t^2)\,dt$, taking the derivative yields $H'(x) = \sin t + t^2$ evaluated at $t=x$. Therefore, $H'(x) = \sin x + x^2$, which is simply the integrand with $t$ replaced by $x$. Choice A ($\cos x + 2x$) might attract students who incorrectly differentiate each term of the integrand rather than evaluating it. To apply FTC Part 1, identify the variable upper limit and substitute it into the integrand—no additional differentiation needed.
9. If $G(x)=\int_{0}^{x}\dfrac{1}{1+t^2}\,dt$, what is $G'(x)$?
   Answer: $\dfrac{1}{1+x^2}$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function G(x) is defined as the integral from 0 to x of 1/(1 + t²) dt, which accumulates the area under the curve of f(t) = 1/(1 + t²) from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative G'(x) equals the integrand evaluated at x, which is 1/(1 + x²). This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice D, which is the integral from 0 to x of 1/(1 + t²) dt, but that actually computes G(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.
10. Suppose $a(x)=\int_{2}^{x}\dfrac{t^2-3}{t^2+3}\,dt$. What is $a'(x)$?
    Answer: $\dfrac{x^2-3}{x^2+3}$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function a(x) is defined as the integral from 2 to x of ($t^2-3$)/($t^2+3$) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative a'(x) is simply the integrand evaluated at t = x, so a'(x) = ($x^2-3$)/($x^2+3$). This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.
11. Define $M(x)=\int_{6}^{x}\dfrac{\sqrt{t}}{1+t}\,dt$ for $x\ge 0$. What is $M'(x)$?
    Answer: $\dfrac{\sqrt{x}}{1+x}$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function M(x) is defined as the integral from 6 to x of $\dfrac{\sqrt{t}}{1+t}\, dt$, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative M'(x) is simply the integrand evaluated at t = x, so M'(x) = $\dfrac{\sqrt{x}}{1+x}$. This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.
12. Let $V(x)=\int_{0}^{x}(\sin t)^2\,dt$. What is $V'(x)$?
    Answer: $\sin^2 x$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function V(x) is defined as the integral from 0 to x of (sin t)² dt, which accumulates the area under the curve of f(t) = (sin t)² from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative V'(x) equals the integrand evaluated at x, which is (sin x)². This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the evaluation of an antiderivative from 0 to x, but that actually computes V(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.
13. If $S(x)=\int_{4}^{x}\dfrac{\cos t}{t}\,dt$, what is $S'(x)$?​
    Answer: $\dfrac{\cos x}{x}$ This problem applies the Fundamental Theorem of Calculus Part 1, which tells us that differentiating an integral with variable upper limit gives the integrand evaluated at that limit. For $S(x) = \int_{4}^{x}\frac{\cos t}{t}\,dt$, taking the derivative yields $S'(x) = \frac{\cos t}{t}$ evaluated at $t=x$. This gives us $S'(x) = \frac{\cos x}{x}$, which is the integrand with $t$ replaced by $x$. Choice B ($\frac{\sin x}{x}$) might attract students who mistakenly think they need to integrate $\cos t$ to get $\sin t$, but we're differentiating the accumulation function, not integrating. To recognize FTC Part 1, look for $\frac{d}{dx}$ of an integral—the answer is always the integrand evaluated at $x$.
14. If $Z(x)=\int_{-1}^{x}\dfrac{1}{(t+2)^2}\,dt$, what is $Z'(x)$?
    Answer: $\dfrac{1}{(x+2)^2}$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function Z(x) is defined as the integral from -1 to x of 1 / (t + 2)² dt, which accumulates the area under the curve of f(t) = 1 / (t + 2)² from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative Z'(x) equals the integrand evaluated at x, which is 1 / (x + 2)². This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the evaluation of an antiderivative from -1 to x, but that actually computes Z(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.
15. Define $H(x)=\int_{-1}^{x}(t-1)^5\,dt$. What is $H'(x)$?
    Answer: $(x-1)^5$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function H(x) is defined as the integral from -1 to x of (t - 1)⁵ dt, which accumulates the area under the curve of f(t) = (t - 1)⁵ from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative H'(x) equals the integrand evaluated at x, which is (x - 1)⁵. This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the evaluation of an antiderivative from -1 to x, but that actually computes H(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.
16. If $F(x)=\int_{-1}^{x}\sqrt{1+t^4}\,dt$, what is $F'(x)$?​​
    Answer: $\sqrt{1+x^4}$ This problem uses the Fundamental Theorem of Calculus Part 1, which tells us that differentiating an accumulation function gives the integrand evaluated at the upper limit. For $F(x) = \int_{-1}^{x}\sqrt{1+t^4}\,dt$, taking the derivative with respect to $x$ yields $F'(x) = \sqrt{1+t^4}$ evaluated at $t=x$. Therefore, $F'(x) = \sqrt{1+x^4}$, which is the integrand with $t$ replaced by $x$. Choice C ($\frac{2x^3}{\sqrt{1+x^4}}$) might appeal to students who mistakenly try to differentiate the integrand expression rather than simply evaluating it. When you see $\frac{d}{dx}\int_{a}^{x}f(t)\,dt$, the answer is always $f(x)$ by FTC Part 1.
17. Let $A(x)=\int_{0}^{x}\big(e^{-t}+t\big)\,dt$. What is $A'(x)$?
    Answer: $e^{-x}+x$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function A(x) is defined as the integral from 0 to x of ($e^{-t} + t$) dt, which accumulates the area under the curve of f(t) = $e^{-t} + t$ from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative A'(x) equals the integrand evaluated at x, which is $e^{-x} + x$. This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the evaluation of an antiderivative from 0 to x, but that actually computes A(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.
18. Let $G(x)=\int_{5}^{x}\dfrac{1}{1+t^2}\,dt$. What is $G'(x)$?​​
    Answer: $\dfrac{1}{1+x^2}$ This is a direct application of the Fundamental Theorem of Calculus Part 1, which states that the derivative of $\int_{a}^{x}f(t)\,dt$ equals $f(x)$. Given $G(x) = \int_{5}^{x}\frac{1}{1+t^2}\,dt$, we find $G'(x)$ by evaluating the integrand at the upper limit $x$. This gives us $G'(x) = \frac{1}{1+x^2}$, obtained by replacing $t$ with $x$ in the integrand. Choice E ($\arctan(x)-\arctan(5)$) might tempt students who recognize that $\int\frac{1}{1+t^2}\,dt = \arctan(t)+C$, but this is the antiderivative, not the derivative we seek. Remember: when differentiating an accumulation function, simply evaluate the integrand at the variable upper limit.
19. Let $E(x)=\int_{-2}^{x}\dfrac{t^2+1}{t^2+4}\,dt$. What is $E'(x)$?
    Answer: $\dfrac{x^2+1}{x^2+4}$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function E(x) is defined as the integral from -2 to x of (t² + 1)/(t² + 4) dt, which accumulates the area under the curve of f(t) = (t² + 1)/(t² + 4) from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative E'(x) equals the integrand evaluated at x, which is (x² + 1)/(x² + 4). This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the integral of (t² + 1)/(t² + 4) dt evaluated from -2 to x, but that actually computes E(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.
20. Let $h(x)=\int_{-6}^{x}\dfrac{1}{t^2+2t+5}\,dt$. What is $h'(x)$?
    Answer: $\dfrac{1}{x^2+2x+5}$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function h(x) is defined as the integral from -6 to x of 1/($t^2+2t+5$) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative h'(x) is simply the integrand evaluated at t = x, so h'(x) = 1/($x^2+2x+5$). This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

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