Fundamental Theorem of Calculus: Accumulation Functions - AP Calculus BC

Let $g(x) = \int_0^x (t^2 - 4t + 3) \, dt$ . For what open intervals is $g(x)$ decreasing?
Answer: $(1, 3)$ A function $g(x)$ is decreasing when its derivative $g'(x)$ is negative. By the Fundamental Theorem of Calculus, $g'(x) = x^2 - 4x + 3$ . To find where $g'(x) < 0$ , we first find the roots: $x^2 - 4x + 3 = (x-1)(x-3) = 0$ , so the roots are $x=1$ and $x=3$ . The parabola $y=x^2 - 4x + 3$ opens upwards, so it is negative between its roots. Thus, $g'(x) < 0$ on the interval $(1, 3)$ .
A tank's net inflow rate is $r(t)=t^2+3t$ . If $A(x)=\int_{1}^{x} r(t)\,dt$ , what is $A'(x)$ ?
Answer: $x^2+3x$ This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1), which states that if A(x) is defined as the integral from a constant to x of r(t) dt, then A'(x) equals r(x). Here, A(x) accumulates the net inflow rate r(t) = t² + 3t from t=1 to t=x, so its derivative A'(x) should simply be the integrand evaluated at x, which is x² + 3x. FTC Part 1 works because the derivative of the accumulated area under the curve r(t) up to x is precisely the height of the function at that point x. This is like how the rate of change of total water in the tank at time x is exactly the inflow rate at that instant. A tempting distractor is choice E, t² + 3t, but it fails because it doesn't replace the dummy variable t with the upper limit x, leaving it in terms of t instead of expressing the derivative as a function of x. To recognize FTC Part 1 problems, look for a function defined as an integral with a variable upper limit and a request for its derivative, which will always be the integrand with t swapped for x.
A thermostat's accumulated deviation is $H(t)=\int_{0}^{t}\sin(x^2)\,dx$ . What is $H'(t)$ ?
Answer: $\sin(t^2)$ This problem demonstrates the Fundamental Theorem of Calculus Part 1 for differentiating an accumulation function. Given $H(t) = \int_{0}^{t}\sin(x^2)\,dx$ , we need to find $H'(t)$ . By FTC Part 1, the derivative of an integral with respect to its upper limit equals the integrand evaluated at that limit. Therefore, $H'(t) = \sin(t^2)$ , obtained by substituting $t$ for $x$ in the integrand. Choice A ( $2t\cos(t^2)$ ) would be correct if we were differentiating $\sin(t^2)$ with respect to $t$ , but FTC Part 1 tells us to simply evaluate the integrand at $t$ . Remember: FTC Part 1 gives us $\frac{d}{dt}\int_{a}^{t} f(x)\,dx = f(t)$ , requiring only substitution.
Let $S(x)=\int_2^x (t^2+1)^{-3}\,dt$ represent accumulated signal. What is $S'(x)$ ?
Answer: $\displaystyle (x^2+1)^{-3}$ This problem illustrates the Fundamental Theorem of Calculus Part 1 for finding the rate of signal accumulation. Since S(x) = ∫₂ˣ (t²+1)⁻³ dt, applying FTC Part 1 gives us S'(x) = (x²+1)⁻³. The theorem states that the derivative of an integral with respect to its upper limit equals the integrand evaluated at that upper limit. We replace every instance of the dummy variable t with x to obtain our answer. Choice C incorrectly keeps the dummy variable, while choice D attempts to differentiate the integrand using the chain rule. To correctly apply FTC Part 1, recognize that x appears only in the limits and substitute it for the dummy variable in the integrand.
A runner's total displacement is $S(x)=\int_{1}^{x}\ln(t)\,dt$ . What is $S'(x)$ ?
Answer: $\ln(x)$ This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1) to find the derivative of an accumulation function. The FTC Part 1 states that if S(x) = ∫ from a constant lower limit to x of f(t) dt, then S'(x) = f(x), provided f is continuous at x. Here, f(t) = ln(t), so S'(x) = ln(x). This equality holds because the derivative of the total displacement is the velocity at x, given by the integrand. A tempting distractor is choice E, which is the integral of 1/t, but while that's related to the antiderivative of ln(t), it doesn't represent the derivative asked for by FTC. To recognize FTC Part 1 problems, look for questions asking for the derivative of a function defined as a definite integral with a variable upper limit.
Let $g(x) = \int_{-2}^x f(t) \, dt$ . What is the value of $g(2)$ ?
Answer: $2\pi$ The value of $g(2)$ is the definite integral $\int_{-2}^2 f(t) \, dt$ . This integral represents the area under the curve of $f(t)$ from $t=-2$ to $t=2$ . The graph described is an upper semicircle of radius 2. The area of a semicircle is $\frac{1}{2}\pi r^2$ . So, the area is $\frac{1}{2}\pi (2)^2 = 2\pi$ .
Let $g(x) = \int_0^x f(t) \, dt$ . The function $f$ is positive and decreasing for $x>0$ . Which of the following statements about the graph of $g$ for $x>0$ must be true?
Answer: It is increasing and concave down. We can determine the properties of $g$ from its derivatives. By the Fundamental Theorem of Calculus, $g'(x) = f(x)$ . We are given that $f$ is positive for $x>0$ , so $g'(x) > 0$ , which means $g$ is increasing. The second derivative is $g''(x) = f'(x)$ . We are given that $f$ is decreasing for $x>0$ , which means $f'(x) < 0$ . Therefore, $g''(x) < 0$ , which means the graph of $g$ is concave down. Thus, the graph of $g$ is increasing and concave down.
A bacteria culture grows at rate $g(t)=e^{2t}$ . If $P(x)=\int_{-1}^{x} g(t)\,dt$ , what is $P'(x)$ ?
Answer: $e^{2x}$ This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1), which states that if $P(x)$ is defined as the integral from a constant to x of $g(t) \, dt$ , then $P'(x)$ equals $g(x)$ . Here, $P(x)$ accumulates the growth rate $g(t) = e^{2t}$ from t=-1 to t=x, so its derivative $P'(x)$ should simply be the integrand evaluated at x, which is $e^{2x}$ . FTC Part 1 works because the derivative of the accumulated population up to x is precisely the growth rate at that point x. This is like how the rate of change of total bacteria at time x is exactly the growth rate at that instant. A tempting distractor is choice B, $e^{2t}$ , but it fails because it doesn't replace the dummy variable t with the upper limit x, leaving it in terms of t instead of expressing the derivative as a function of x. To recognize FTC Part 1 problems, look for a function defined as an integral with a variable upper limit and a request for its derivative, which will always be the integrand with t swapped for x.
A lake's pollution changes at rate $p(t)=t\cos t$ . If $M(x)=\int_{3}^{x} p(t)\,dt$ , what is $M'(x)$ ?
Answer: $x\cos x$ This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1), which states that if M(x) is defined as the integral from a constant to x of p(t) dt, then M'(x) equals p(x). Here, M(x) accumulates the pollution rate p(t) = t cos t from t=3 to t=x, so its derivative M'(x) should simply be the integrand evaluated at x, which is x cos x. FTC Part 1 works because the derivative of the accumulated pollution up to x is precisely the change rate at that point x. This is like how the rate of change of total pollution at time x is exactly the pollution rate at that instant. A tempting distractor is choice B, t cos t, but it fails because it doesn't replace the dummy variable t with the upper limit x, leaving it in terms of t instead of expressing the derivative as a function of x. To recognize FTC Part 1 problems, look for a function defined as an integral with a variable upper limit and a request for its derivative, which will always be the integrand with t swapped for x.
Define $P(x)=\int_{3}^{x}\frac{t}{\sqrt{t^2+9}}\,dt$ as accumulated work. What is $P'(x)$ ?
Answer: $\frac{x}{\sqrt{x^2+9}}$ This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1) to find the derivative of an accumulation function. The FTC Part 1 states that if P(x) = ∫ from a constant lower limit to x of f(t) dt, then P'(x) = f(x), provided f is continuous at x. For this, f(t) = t / √(t² + 9), so P'(x) = x / √(x² + 9). This principle shows the derivative is the integrand substituted with x. A tempting distractor is choice E, which integrates an expression resembling the derivative of the integrand, but that confuses FTC with unnecessary differentiation. To recognize FTC Part 1 problems, look for questions asking for the derivative of a function defined as a definite integral with a variable upper limit.
Let $g(x) = \frac{d}{dx} \int_1^{\ln x} (t^2 + e^t) \, dt$ . Find $g(e)$ .
Answer: $\frac{1+e}{e}$ First, find the expression for $g(x)$ . Using the Fundamental Theorem of Calculus with the chain rule, $g(x) = ((\ln x)^2 + e^{\ln x}) \cdot \frac{d}{dx}(\ln x) = ((\ln x)^2 + x) \cdot \frac{1}{x}$ . Now, evaluate $g(e)$ : $g(e) = ((\ln e)^2 + e) \cdot \frac{1}{e} = (1^2 + e) \cdot \frac{1}{e} = \frac{1+e}{e}$ .
Let $F(x) = \int_0^x (t-1)^2 (t-5) \, dt$ . At what value of $x$ does $F$ have a local minimum?
Answer: $x=5$ To find local extrema of $F$ , we examine its derivative, $F'(x)$ . By the Fundamental Theorem of Calculus, $F'(x) = (x-1)^2(x-5)$ . Critical points occur when $F'(x)=0$ , which are at $x=1$ and $x=5$ . For a local minimum, $F'(x)$ must change from negative to positive. The term $(x-1)^2$ is always non-negative. The sign of $F'(x)$ is determined by the term $(x-5)$ . For $x<5$ , $F'(x) < 0$ . For $x>5$ , $F'(x) > 0$ . Therefore, a local minimum occurs at $x=5$ . At $x=1$ , the derivative does not change sign.
If $H(x) = \int_{\sin(x)}^1 e^{t^2} \, dt$ , find $H'(x)$ .
Answer: $-\cos(x) e^{\sin^2(x)}$ First, reverse the limits of integration: $H(x) = -\int_1^{\sin(x)} e^{t^2} \, dt$ . Now, apply the Fundamental Theorem of Calculus with the chain rule. Let $u(x) = \sin(x)$ , so $u'(x) = \cos(x)$ . Then $H'(x) = - (e^{(\sin(x))^2} \cdot \cos(x)) = -\cos(x) e^{\sin^2(x)}$ .
Let $F(x) = \int_{2x}^{x^3} \frac{1}{1+t^2} \, dt$ . What is $F'(x)$ ?
Answer: $\frac{3x^2}{1+x^6} - \frac{2}{1+4x^2}$ When both limits of integration are functions of x, we use the rule $\frac{d}{dx} \int_{v(x)}^{u(x)} f(t) \, dt = f(u(x))u'(x) - f(v(x))v'(x)$ . Here, $f(t) = \frac{1}{1+t^2}$ , $u(x) = x^3$ , and $v(x) = 2x$ . So, $u'(x) = 3x^2$ and $v'(x) = 2$ . Applying the rule, $F'(x) = \frac{1}{1+(x^3)^2} \cdot 3x^2 - \frac{1}{1+(2x)^2} \cdot 2 = \frac{3x^2}{1+x^6} - \frac{2}{1+4x^2}$ .
Let $g(x) = \int_a^x f(t) \, dt$ and $h(x) = \int_b^x f(t) \, dt$ , where $a$ and $b$ are constants. Which of the following statements must be true for all $x$ in the domain of $f$ ?
Answer: The graphs of $g$ and $h$ are parallel curves. By the Fundamental Theorem of Calculus, $g'(x) = f(x)$ and $h'(x) = f(x)$ . Since their derivatives are equal, the graphs of $g(x)$ and $h(x)$ are parallel curves; they differ by a constant. Specifically, $g(x) - h(x) = \int_a^x f(t)dt - \int_b^x f(t)dt = \int_a^b f(t)dt$ , which is a constant. This means the graph of $g(x)$ is a vertical translation of the graph of $h(x)$ .
Let $H(x) = \int_2^x \sqrt{1+t^3} \, dt$ . An equation for the line tangent to the graph of $H$ at $x=2$ is
Answer: $y = 3(x-2)$ The equation of a tangent line is $y - y_1 = m(x-x_1)$ . The point of tangency is $(x_1, y_1) = (2, H(2))$ . $H(2) = \int_2^2 \sqrt{1+t^3} \, dt = 0$ . So the point is $(2,0)$ . The slope $m$ is $H'(2)$ . By the Fundamental Theorem of Calculus, $H'(x) = \sqrt{1+x^3}$ . So, $m = H'(2) = \sqrt{1+2^3} = \sqrt{9} = 3$ . The equation of the tangent line is $y - 0 = 3(x-2)$ , or $y = 3(x-2)$ .
A chemical's total reacted amount is $R(t)=\int_{0}^{t}e^{-0.2x}\,dx$ . What is $R'(t)$ ?
Answer: $e^{-0.2t}$ This problem requires applying the Fundamental Theorem of Calculus Part 1 to differentiate an accumulation function. Given $R(t) = \int_{0}^{t}e^{-0.2x}\,dx$ , we need to find $R'(t)$ . By FTC Part 1, the derivative of an integral with respect to its upper limit equals the integrand evaluated at that upper limit. Therefore, $R'(t) = e^{-0.2t}$ , obtained by substituting $t$ for $x$ in the integrand. Choice A represents the derivative of $e^{-0.2t}$ , which students might choose if they mistakenly differentiate the result, but FTC Part 1 requires only evaluation at $t$ . To recognize FTC Part 1 problems, look for derivatives of integrals where the variable appears only in the limits.
Let $f(x) = \begin{cases} 3 & \text{for } 0 \le x < 1 \\ 3x^2 & \text{for } x \ge 1 \end{cases}$ . Let $g(x) = \int_0^x f(t) \, dt$ . What is $g(2)$ ?
Answer: 10 To find $g(2)$ , we must split the integral at $x=1$ because the definition of $f(x)$ changes there. $g(2) = \int_0^2 f(t) \, dt = \int_0^1 f(t) \, dt + \int_1^2 f(t) \, dt$ . For the first integral, $f(t)=3$ , and for the second, $f(t)=3t^2$ . So, $g(2) = \int_0^1 3 \, dt + \int_1^2 3t^2 \, dt = [3t]_0^1 + [t^3]_1^2 = (3(1)-3(0)) + (2^3 - 1^3) = 3 + (8-1) = 3+7=10$ .
A particle's velocity is $v(t)=t^2\sin t$ . If $A(x)=\int_0^x v(t)\,dt$ , what is $A'(x)$ ?
Answer: $\displaystyle x^2\sin x$ This problem requires applying the Fundamental Theorem of Calculus Part 1, which states that the derivative of an accumulation function equals the integrand evaluated at the upper limit. Since A(x) = ∫₀ˣ v(t) dt where v(t) = t²sin t, we have A'(x) = v(x) = x²sin x. The FTC Part 1 tells us that when we differentiate an integral with respect to its upper limit, we simply substitute that upper limit into the integrand. Choice A incorrectly suggests the derivative is the integral itself, failing to apply the FTC. To recognize FTC Part 1 problems, look for derivatives of integrals where the variable appears only in the limits of integration.
A heater adds energy at rate $h(t)=\frac{1}{1+t^2}$ . Let $E(x)=\int_{0}^{x} h(t)\,dt$ . What is $E'(x)$ ?
Answer: $\frac{1}{1+x^2}$ This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1), which states that if E(x) is defined as the integral from a constant to x of h(t) dt, then E'(x) equals h(x). Here, E(x) accumulates the energy rate h(t) = 1/(1+t²) from t=0 to t=x, so its derivative E'(x) should simply be the integrand evaluated at x, which is 1/(1+x²). FTC Part 1 works because the derivative of the accumulated energy up to x is precisely the heating rate at that point x. This is like how the rate of change of total energy at time x is exactly the input rate at that instant. A tempting distractor is choice B, 1/(1+t²), but it fails because it doesn't replace the dummy variable t with the upper limit x, leaving it in terms of t instead of expressing the derivative as a function of x. To recognize FTC Part 1 problems, look for a function defined as an integral with a variable upper limit and a request for its derivative, which will always be the integrand with t swapped for x.

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