📝 Differentiating integrals with variable limits (24 MCQs)

📖 From Calculus • 6. Integration • 24 questions available

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What is Differentiating integrals with variable limits?

Definition:
If $F(x) = \int_{a(x)}^{b(x)} f(t) \, dt$ , then $F'(x) = f(b(x))b'(x) - f(a(x))a'(x)$ . This generalizes FTC Part 1 using the Chain Rule for both upper and lower limits.

Example:
$F(x) = \int_x^{x^2} e^t dt$ . Derivative: $e^{x^2}(2x) - e^x(1) = 2xe^{x^2} - e^x$ .

Reason:
This rule is crucial for problems where integration bounds depend on the variable, common in probability theory and advanced physics applications.

Easy

Medium

Hard

📝 All Differentiating integrals with variable limits MCQs

Q1. A student computes the derivative of $F(x) = \int_{0}^{x^2} e^{t^2} \, dt$ as $e^{x^2}$ . What error did the student make?

A.They forgot to evaluate the integrand at the upper limit.

B.They applied the chain rule incorrectly by not multiplying by the derivative of

x^2

. ✅

C.They integrated

e^{t^2}

incorrectly.

D.They used the lower limit instead of the upper limit.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The student correctly identified that the derivative of an integral with a variable upper limit is the integrand evaluated at that limit, but they forgot the chain rule. The correct derivative is $2x e^{x^4}$ , as per the formula $F'(x) = f(g(x))g'(x)$ . This is a classic and common error, making it a good conceptual check for the chain rule's Easy in this context.

Q2. Find the derivative of $F(x) = \int_{x}^{1} \cos(t^2) \, dt$ .

-\cos(x^2)

✅

\cos(x^2)

2x \cos(x^2)

-2x \cos(x^2)

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: To differentiate an integral with a variable lower limit and a constant upper limit, we use the property $\frac{d}{dx} \int_{g(x)}^{a} f(t)dt = -f(g(x))g'(x)$ . Applying this, $F'(x) = -\cos(x^2) \cdot 1 = -\cos(x^2)$ . This question tests the understanding of how to handle limits and the negative sign that comes from reversing the limits of integration.

Q3. If $F(x) = \int_{2x}^{x^2} \sqrt{t^3 + 1} \, dt$ , what is $F'(x)$ ?

2x\sqrt{x^6 + 1} - 2\sqrt{8x^3 + 1}

\sqrt{x^6 + 1} - 2\sqrt{8x^3 + 1}

2x\sqrt{x^6 + 1} - 2\sqrt{8x^3 + 1}

✅

2x\sqrt{x^6 + 1} - 2\sqrt{8x^3 + 1}

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: This problem requires applying the formula $\frac{d}{dx} \int_{g(x)}^{h(x)} f(t)dt = f(h(x))h'(x) - f(g(x))g'(x)$ . Substituting $f(t) = \sqrt{t^3 + 1}$ , $h(x)=x^2$ , and $g(x)=2x$ , we get $F'(x) = \sqrt{(x^2)^3 + 1} \cdot 2x - \sqrt{(2x)^3 + 1} \cdot 2$ . This simplifies to $2x\sqrt{x^6 + 1} - 2\sqrt{8x^3 + 1}$ .

Q4. What is the derivative of $F(x) = \int_{0}^{\sin x} \ln(1+t) \, dt$ at $x = \pi/2$ ?

0

1

\ln 2

✅

\ln(1+\sin x)

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The derivative is $F'(x) = \ln(1+\sin x) \cdot \cos x$ . At $x = \pi/2$ , this becomes $\ln(1+1) \cdot \cos(\pi/2) = \ln 2 \cdot 0 = 0$ . The common mistake is to stop at $\ln(1+\sin x)$ , but the chain rule factor $\cos x$ makes the final value zero at this specific point, testing careful evaluation.

Q5. The graph of $f(t)$ is a semicircle of radius 2 above the t-axis from $t=-2$ to $t=2$ . If $F(x) = \int_{0}^{x} f(t)dt$ , what is the best description of $F'(x)$ on the interval (0,2)?

F&#039;(x)

is positive and decreasing. ✅

F&#039;(x)

is positive and increasing.

F&#039;(x)

is negative and decreasing.

F&#039;(x)

is negative and increasing.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: By the Fundamental Theorem of Calculus, $F'(x) = f(x)$ . The graph of $f$ is the upper semicircle, so on (0,2), $f(x)$ is positive but decreasing (the curve slopes downward toward zero). This question moves from symbolic to graphical interpretation, requiring the student to visualize the function and describe its behavior on an interval.

Q6. Let $f$ be a continuous, even function. For $F(x) = \int_{0}^{x} f(t)dt$ , which statement is true?

F(x)

is an even function.

F(x)

is an odd function. ✅

F(x)

is always increasing.

F(x)

is always positive.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: If $f$ is even ( $f(-t)=f(t)$ ), then $F(-x) = \int_{0}^{-x} f(t)dt = -\int_{0}^{x} f(-u)du = -\int_{0}^{x} f(u)du = -F(x)$ . Thus, $F$ is odd. This tests the interplay between function properties (even/odd) and integration, a higher-level conceptual link.

Q7. Find $\frac{d}{dx} \left[ \int_{\ln x}^{x} \frac{dt}{t} \right]$ .

\frac{1}{x} - \frac{1}{\ln x} \cdot \frac{1}{x}

\frac{1}{x} - \frac{1}{\ln x}

\frac{1}{x} - \frac{1}{x \ln x}

✅

\frac{1}{x} + \frac{1}{x \ln x}

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Using the formula $f(h(x))h'(x) - f(g(x))g'(x)$ with $f(t)=1/t$ , $h(x)=x$ , $g(x)=\ln x$ , we get $F'(x) = \frac{1}{x} \cdot 1 - \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x} - \frac{1}{x \ln x}$ . This combines logarithmic differentiation with the integral differentiation rule.

Q8. Given $F(x) = \int_{1}^{x^2} f(t)dt$ , which statement about $F''(x)$ is correct?

F&#039;&#039;(x) = f&#039;(x^2) \cdot 2x

F&#039;&#039;(x) = f(x^2) \cdot 2 + 2x f&#039;(x^2) \cdot 2x

F&#039;&#039;(x) = 2x f&#039;(x^2)

F&#039;&#039;(x) = 2 f(x^2) + 4x^2 f&#039;(x^2)

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: First, $F'(x) = 2x f(x^2)$ . Then, to find $F''(x)$ , we must use the product rule and the chain rule: $F''(x) = 2f(x^2) + 2x f'(x^2) \cdot 2x = 2f(x^2) + 4x^2 f'(x^2)$ . This is a multi-step problem testing differentiation rules beyond the first derivative.

Q9. A student claims $\frac{d}{dx} \int_{0}^{x} \sin(t^2)dt = \sin(x^2)$ . Is the student correct, and if not, why?

A.Yes, the student is correct. ✅

B.No, the derivative is

2x \cos(x^2)

C.No, the derivative is

\cos(x^2)

D.No, the derivative is

2x \sin(x^2)

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The student is absolutely correct. The Fundamental Theorem of Calculus Part 2 states that $F'(x) = f(x)$ for $F(x) = \int_{a}^{x} f(t)dt$ . The chain rule is not needed here because the upper limit is just $x$ , not a function of $x$ . This problem tests the ability to distinguish between the simple and chain-rule cases.

Q10. Find $\lim_{h \to 0} \frac{1}{h} \int_{2}^{2+h} e^{-t^2} dt$ .

e^{-4}

✅

2e^{-4}

0

e^{-4}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This limit is, by definition, the derivative of $F(x) = \int_{2}^{x} e^{-t^2}dt$ at $x=2$ . By the Fundamental Theorem of Calculus, $F'(2) = e^{-2^2} = e^{-4}$ . This requires recognizing the limit definition of the derivative within the context of integrals, connecting limit and integral concepts.

Q11. Find the derivative of $F(x) = \int_{x^2}^{x^3} \frac{1}{1+t^2} dt$ .

\frac{3x^2}{1+x^6} - \frac{2x}{1+x^4}

✅

\frac{1}{1+x^6} - \frac{1}{1+x^4}

\frac{3x^2}{1+x^6} + \frac{2x}{1+x^4}

\frac{3x^2}{1+x^6} - \frac{2x}{1+x^4}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Applying the formula $\frac{d}{dx} \int_{g(x)}^{h(x)} f(t)dt = f(h(x))h'(x) - f(g(x))g'(x)$ , we get $\frac{1}{1+(x^3)^2} \cdot 3x^2 - \frac{1}{1+(x^2)^2} \cdot 2x = \frac{3x^2}{1+x^6} - \frac{2x}{1+x^4}$ . This is a direct, one-step Easy of the standard formula.

Q12. Let $F(x) = \int_{0}^{x^2} \sqrt{1+t^3} dt$ . What is $F'(1)$ ?

2

\sqrt{2}

2\sqrt{2}

✅

3\sqrt{2}

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Here, $F'(x) = 2x\sqrt{1+(x^2)^3} = 2x\sqrt{1+x^6}$ . Evaluating at $x=1$ : $F'(1) = 2(1)\sqrt{1+1} = 2\sqrt{2}$ . This combines the derivative rule with evaluation, a standard but essential multi-step process.

Q13. Given $F(x) = \int_{-x}^{x} \sin(t^2) dt$ , find $F'(x)$ .

2x \sin(x^2)

2 \sin(x^2)

✅

0

\sin(x^2) - \sin(x^2) = 0

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Using the property $\frac{d}{dx} \int_{-x}^{x} f(t)dt = f(x) \cdot 1 - f(-x) \cdot (-1) = f(x) + f(-x)$ . Since $f(t)=\sin(t^2)$ is even, $f(x)+f(-x)=2f(x)=2\sin(x^2)$ . This question combines the derivative rule with the property of even functions.

Q14. A particle's position is given by $s(t) = \int_{0}^{t} \cos(u^2) du$ . What is its acceleration?

-\sin(t^2)

-\sin(t^2) \cdot 2t

✅

\cos(t^2)

-2t \sin(t^2)

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Velocity is $s'(t) = \cos(t^2)$ . Acceleration is the derivative of velocity: $a(t) = -\sin(t^2) \cdot 2t = -2t \sin(t^2)$ . This frames the mathematical concept within a physics model, requiring an extra step of differentiation after finding the velocity.

Q15. For $F(x) = \int_{\sqrt{x}}^{x} \cos(t^2) dt$ , which of the following is NOT a correct step in finding the derivative?

F&#039;(x) = \cos(x^2) - \cos(x) \cdot \frac{1}{2\sqrt{x}}

✅

F&#039;(x) = \cos(x^2) - \frac{1}{2\sqrt{x}} \cos(x)

F&#039;(x) = \cos(x^2) - \frac{\cos(\sqrt{x}^2)}{2\sqrt{x}}

F&#039;(x) = \cos(x^2) - \frac{\cos(x)}{2\sqrt{x}}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The correct derivative is $\cos(x^2) \cdot 1 - \cos((\sqrt{x})^2) \cdot \frac{1}{2\sqrt{x}} = \cos(x^2) - \frac{\cos(x)}{2\sqrt{x}}$ . Option A incorrectly uses $\cos(x)$ instead of $\cos(x) \cdot \frac{1}{2\sqrt{x}}$ , missing the chain rule factor for the lower limit.

Q16. Find $\frac{d}{dx} \left[ \int_{1}^{e^x} \ln t \, dt \right]$ .

e^x

\ln(e^x)

x e^x

\ln(e^x) \cdot e^x = x e^x

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Let $F(x) = \int_{1}^{e^x} \ln t \, dt$ . Then $F'(x) = \ln(e^x) \cdot e^x = x e^x$ . This tests the chain rule with an exponential upper limit, requiring careful Easy of both parts.

Q17. If $F(x) = \int_{0}^{x} f(t)dt$ and the graph of $f$ is a straight line with negative slope from (0,2) to (2,0), on which interval is $F(x)$ concave down?

A.-1

B.-12

C.-2 ✅

F

is never concave down.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: $F''(x) = f'(x)$ . Since the graph of $f$ is a line with a negative slope over the entire interval (0,2), $f'(x)$ is negative for all $x$ in (0,2). Therefore, $F$ is concave down on the entire interval (0,2).

Q18. A function $F(x) = \int_{0}^{x} f(t)dt$ is given. If $F(2)=5$ and $F'(x)=x^2$ , what is the value of $\int_{0}^{2} f(t)dt$ ?

A.5 ✅

\frac{8}{3}

\frac{8}{3} + 5

D.Cannot be determined.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: By the Fundamental Theorem of Calculus, $\int_{0}^{2} f(t)dt = F(2) - F(0)$ . Since $F(0) = \int_{0}^{0} f(t)dt = 0$ , the integral equals $F(2) = 5$ . The given derivative is a distractor; the value is directly given by the property of the integral.

Q19. Find the derivative of $F(x) = \int_{x}^{x^2} \frac{dt}{\sqrt{1+t^2}}$ .

\frac{2x}{\sqrt{1+x^4}} - \frac{1}{\sqrt{1+x^2}}

✅

\frac{1}{\sqrt{1+x^4}} - \frac{1}{\sqrt{1+x^2}}

\frac{2x}{\sqrt{1+x^4}} + \frac{1}{\sqrt{1+x^2}}

\frac{2x}{\sqrt{1+x^4}} - \frac{1}{\sqrt{1+x^2}}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Applying the general formula: $F'(x) = f(x^2) \cdot 2x - f(x) \cdot 1 = \frac{2x}{\sqrt{1+x^4}} - \frac{1}{\sqrt{1+x^2}}$ . This is a straightforward Easy of the formula.

Q20. Let $F(x) = \int_{\cos x}^{\sin x} e^{t^2} dt$ . What is $F'(\pi/4)$ ?

e^{1/2} \cdot \frac{\sqrt{2}}{2} + e^{1/2} \cdot \frac{\sqrt{2}}{2} = \sqrt{2e}

e^{1/2} \cdot \frac{\sqrt{2}}{2} - e^{1/2} \cdot \frac{\sqrt{2}}{2} = 0

✅

e^{1/2} \cdot \frac{\sqrt{2}}{2} + e^{1/2} \cdot \frac{\sqrt{2}}{2} = \sqrt{2} e^{1/2}

e^{1/2} \cdot \frac{\sqrt{2}}{2} - e^{1/2} \cdot \frac{\sqrt{2}}{2} = 0

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: $F'(x) = e^{(\sin x)^2} \cos x - e^{(\cos x)^2} (-\sin x) = e^{\sin^2 x} \cos x + e^{\cos^2 x} \sin x$ . At $x=\pi/4$ , $\sin x = \cos x = \sqrt{2}/2$ , so $F'(\pi/4) = e^{1/2} \cdot \frac{\sqrt{2}}{2} + e^{1/2} \cdot \frac{\sqrt{2}}{2} = \sqrt{2} e^{1/2}$ . This requires evaluating trig functions and simplifying.

Q21. If $F(x) = \int_{1}^{2x} t^3 dt$ , which is larger, $F'(1)$ or $F(1)$ ?

F&#039;(1)

is larger. ✅

F(1)

is larger.

C.They are equal.

D.Cannot be determined.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: $F(x) = \left[ \frac{t^4}{4} \right]_{1}^{2x} = \frac{16x^4}{4} - \frac{1}{4} = 4x^4 - \frac{1}{4}$ . So $F(1) = 4 - 0.25 = 3.75$ . Also, $F'(x) = 16x^3$ , so $F'(1) = 16$ . Since 16 > 3.75, $F'(1)$ is larger. This combines integration, differentiation, and comparison.

Q22. What is $\frac{d}{dx} \left[ \int_{x}^{x^2} \sqrt{1+t^4} dt \right]$ at $x = 1$ ?

A.0

2\sqrt{2}

2\sqrt{2} - \sqrt{2} = \sqrt{2}

✅

\sqrt{2}

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: $F'(x) = \sqrt{1+x^8} \cdot 2x - \sqrt{1+x^4} \cdot 1$ . At $x=1$ : $F'(1) = \sqrt{2} \cdot 2 - \sqrt{2} \cdot 1 = \sqrt{2}$ . This is a direct Easy but requires careful evaluation and simplification at the specific point.

Q23. If $F(x) = \int_{0}^{x^2} \sin(t^2) dt$ , then $F'''(0)$ is:

A.0 ✅

B.2

C.-2

D.4

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: $F'(x) = 2x \sin(x^4)$ . $F''(x) = 2\sin(x^4) + 8x^4 \cos(x^4)$ . $F'''(x) = 8x^3 \cos(x^4) + 32x^3 \cos(x^4) - 32x^7 \sin(x^4) = 40x^3 \cos(x^4) - 32x^7 \sin(x^4)$ . Evaluating at $x=0$ gives 0. This is a complex, multi-step differentiation problem testing higher-order derivatives.

Q24. A student tries to evaluate $\lim_{x \to 0} \frac{\int_{0}^{x^2} \cos(t) dt}{x^2}$ using L'Hôpital's rule. What is the result?

\frac{1}{2}

B.1 ✅

C.0

D.The limit does not exist.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The numerator is $F(x) = \int_{0}^{x^2} \cos t dt$ . As $x \to 0$ , both numerator and denominator tend to 0. Applying L'Hôpital's rule gives $\lim_{x \to 0} \frac{F'(x)}{2x} = \lim_{x \to 0} \frac{2x \cos(x^2)}{2x} = \lim_{x \to 0} \cos(x^2) = 1$ . This problem tests the correct Easy of L'Hôpital's rule combined with the integral derivative.

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