Calculus 2 Quiz: Ftc And Definite Integrals

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Ftc And Definite IntegralsQuestion 1 of 20

If $h(x)$ is continuous and $\int_1^5 h(x) dx = 8$ , then $\lim_{n \to \infty} \frac{4}{n} \sum_{k=1}^{n} h\left(1 + \frac{4k}{n}\right)$ equals:

8

2

16

32

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Calculus 2 Quiz

Calculus 2 Quiz: Ftc And Definite Integrals

Practice Ftc And Definite Integrals in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Ftc And Definite Integrals, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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Question 1

If $h(x)$ is continuous and $\int_1^5 h(x) dx = 8$ , then $\lim_{n \to \infty} \frac{4}{n} \sum_{k=1}^{n} h\left(1 + \frac{4k}{n}\right)$ equals:

$8$ (correct answer)
$2$
$16$
$32$

Explanation: When you see a limit involving a sum with terms like

\frac{1}{n}\sum_{k=1}^{n} f\left(a + \frac{(b-a)k}{n}\right)

, you're looking at a Riemann sum that approaches a definite integral as

n \to \infty

. Let's identify the components in

\lim_{n \to \infty} \frac{4}{n} \sum_{k=1}^{n} h\left(1 + \frac{4k}{n}\right)

. Here,

\Delta x = \frac{4}{n}

, the lower bound is

a = 1

, and the upper bound is

b = 1 + 4 = 5

. The expression

1 + \frac{4k}{n}

represents sample points in the interval

[1,5]

using right endpoints. By the definition of definite integrals as limits of Riemann sums, this limit equals:

\int_1^5 h(x) dx

Since we're given that

\int_1^5 h(x) dx = 8

, the answer is A) 8. Let's examine why the other choices are wrong:

B) 2 might result from incorrectly dividing the integral by the width of the interval ( $8 ÷ 4 = 2$ ), but Riemann sums converge to the integral itself, not some scaled version.
C) 16 could come from mistakenly multiplying the given integral by 2, perhaps confusing the coefficient 4 in the sum.
D) 32 likely results from multiplying by 4, incorrectly thinking the factor $\frac{4}{n}$ somehow multiplies the final result.

Study tip: When you see Riemann sum limits, identify the interval endpoints and recognize that the limit always equals the definite integral over that interval—no additional scaling needed.

Question 2

Let $f(x) = \int_2^x (t^2 - 4t + 3) dt$ . If $f'(c) = 0$ for some $c > 2$ , what is the value of $c$ ?

$c = 1$
$c = 3$ (correct answer)
$c = 4$
$c = 5$

Explanation: By the Fundamental Theorem of Calculus,

f'(x) = x^2 - 4x + 3

. Setting

f'(c) = 0

c^2 - 4c + 3 = 0

, which factors as

(c-1)(c-3) = 0

. This gives

c = 1

c = 3

. Since we need

c > 2

, we have

c = 3

. Choice A gives

c = 1

, which doesn't satisfy

c > 2

. Choices C and D are incorrect factorizations or solutions to the quadratic equation.

Question 3

Let $F(x) = \int_{x^2}^{2x} \sin(t^3) dt$ . Using the Fundamental Theorem of Calculus, $F'(x)$ equals:

$\sin((2x)^3) - \sin((x^2)^3)$
$2\sin((2x)^3) - 2x\sin((x^2)^3)$
$\sin(8x^3) - \sin(x^6)$
$2\sin(8x^3) - 2x\sin(x^6)$ (correct answer)

Explanation: Using the chain rule with FTC:

F'(x) = \sin((2x)^3) \cdot \frac{d}{dx}(2x) - \sin((x^2)^3) \cdot \frac{d}{dx}(x^2) = \sin(8x^3) \cdot 2 - \sin(x^6) \cdot 2x = 2\sin(8x^3) - 2x\sin(x^6)

. Choice A omits the chain rule derivatives. Choice B has the correct derivatives but wrong simplification of the cubic terms. Choice C has correct simplification but omits the chain rule derivatives.

Question 4

Let $f$ be a continuous function. Given $\int_1^7 f(x) dx = 10$ and $\int_5^7 (f(x) - 3) dx = 2$ , what is the value of $\int_1^5 f(x) dx$ ?

$2$ (correct answer)
$5$
$8$
$12$

Explanation: This problem requires using the properties of definite integrals. First, simplify the second given integral:

\int_5^7 (f(x) - 3) dx = \int_5^7 f(x) dx - \int_5^7 3 dx = 2

. The second term is

3(7-5) = 3(2) = 6

. So,

\int_5^7 f(x) dx - 6 = 2

, which implies

\int_5^7 f(x) dx = 8

. Now use the additivity property of integrals:

\int_1^7 f(x) dx = \int_1^5 f(x) dx + \int_5^7 f(x) dx

. Substituting the known values:

10 = \int_1^5 f(x) dx + 8

. Solving for the unknown integral gives

\int_1^5 f(x) dx = 10 - 8 = 2

Question 5

Given that $\int_0^x f(t) dt = 3x^2 + \sin(\pi x)$ , what is the value of $f(2)$ ?

$12$
$12 + \pi$ (correct answer)
$12 - \pi$
$6$

Explanation: This problem uses the Fundamental Theorem of Calculus, Part 1. We are given an equation relating a function

f

to its integral. To find

f(x)

, we differentiate both sides of the equation with respect to

x

\frac{d}{dx} \int_0^x f(t) dt = \frac{d}{dx} (3x^2 + \sin(\pi x))

. The left side becomes

f(x)

by FTC1. The derivative of the right side is

6x + \cos(\pi x) \cdot \pi

. So,

f(x) = 6x + \pi \cos(\pi x)

. To find

f(2)

, we substitute

x=2

f(2) = 6(2) + \pi \cos(2\pi) = 12 + \pi(1) = 12 + \pi

Question 6

Evaluate the integral $\int_e^{e^4} \frac{dx}{x \sqrt{\ln x}}$ .

$1$
$3$
$3/2$
$2$ (correct answer)

Explanation: When you encounter an integral with both

x

and

\ln x

in the expression, substitution is typically your best approach. The key insight here is recognizing that the derivative of

\ln x

\frac{1}{x}

, which appears as a factor in our integrand. Let's use the substitution

u = \ln x

, so

du = \frac{1}{x}dx

. We need to change our limits of integration: when

x = e

, we have

u = \ln e = 1

, and when

x = e^4

, we have

u = \ln(e^4) = 4

. Our integral becomes:

\int_1^4 \frac{du}{\sqrt{u}} = \int_1^4 u^{-1/2} du

Using the power rule for integration:

\int u^{-1/2} du = \frac{u^{1/2}}{1/2} = 2\sqrt{u}

Evaluating from 1 to 4:

2\sqrt{u}\Big|_1^4 = 2\sqrt{4} - 2\sqrt{1} = 2(2) - 2(1) = 4 - 2 = 2

The answer is D)

2

. Choice A)

1

likely results from forgetting the factor of 2 in the antiderivative. Choice B)

3

might come from incorrectly evaluating

\sqrt{4} - \sqrt{1}

without the coefficient. Choice C)

3/2

could result from using the wrong power rule or making an arithmetic error with the coefficient. Study tip: When you see

\ln x

in an integral, always consider

u = \ln x

substitution, especially when

\frac{1}{x}

appears elsewhere in the integrand. This pattern shows up frequently in Calculus 2.

Question 7

If $f(x)$ is a continuous odd function and $\int_{-2}^5 f(x) dx = 8$ , what is the value of $\int_2^5 f(x) dx$ ?

$0$
$4$
$8$ (correct answer)
$16$

Explanation: We can split the given integral using the additivity property:

\int_{-2}^5 f(x) dx = \int_{-2}^2 f(x) dx + \int_2^5 f(x) dx

. Since

f(x)

is an odd function, its integral over a symmetric interval

[-a, a]

is zero. Therefore,

\int_{-2}^2 f(x) dx = 0

. Substituting this into our equation gives

8 = 0 + \int_2^5 f(x) dx

. Thus,

\int_2^5 f(x) dx = 8

Question 8

A student attempts to evaluate $\int_{-2}^2 \frac{1}{x^4} dx$ and finds $[-\frac{1}{3x^3}]_{-2}^2 = (-\frac{1}{24}) - (-\frac{1}{-24}) = -\frac{1}{12}$ . Which of the following best explains the student's error?

The antiderivative was calculated incorrectly; it should be $-\frac{3}{x^5}$ .
The Fundamental Theorem of Calculus is not applicable because the integrand is discontinuous on the interval $[-2, 2]$ . (correct answer)
The integrand is an even function, so the integral should be $2 \int_{0}^2 \frac{1}{x^4} dx$ , which simplifies the calculation.
A sign error was made during the evaluation of the antiderivative at the limits of integration.

Explanation: The Fundamental Theorem of Calculus (Part 2) requires the integrand to be continuous on the closed interval of integration. The function

f(x) = \frac{1}{x^4}

has an infinite discontinuity at

x=0

, which is within the interval

[-2, 2]

. Therefore, the theorem cannot be applied directly. This is an improper integral which, in this case, diverges. The student's calculation, while mechanically following the FTC steps, leads to a meaningless result.

Question 9

Which of the following inequalities is correct without explicit calculation?

$\int_1^2 \ln(x) dx > \int_1^2 \ln(x^2) dx$
$\int_0^{\pi/2} \sin(x) dx > \int_0^{\pi/2} x dx$
$\int_0^1 e^{-x} dx < \int_0^1 e^{-x^2} dx$ (correct answer)
$\int_2^3 \frac{1}{x} dx > \int_2^3 \frac{1}{\sqrt{x}} dx$

Explanation: This question uses the comparison property of integrals: if

f(x) \ge g(x)

[a, b]

, then

\int_a^b f(x) dx \ge \int_a^b g(x) dx

. For option C: On the interval

(0, 1)

, we have

0 < x^2 < x < 1

, which implies

-x^2 > -x

. Since the exponential function is strictly increasing,

e^{-x^2} > e^{-x}

for all

x \in (0, 1)

. By the comparison property,

\int_0^1 e^{-x^2} dx > \int_0^1 e^{-x} dx

, making option C correct. The other options can be shown to be false by similar comparison arguments.

Question 10

Consider the function $G(x) = \int_{\cos x}^{\sin x} e^{t^2} dt$ . The value $G'\left(\frac{\pi}{4}\right)$ equals:

$\frac{\sqrt{2}}{2}(e^{1/2} - e^{1/2}) = 0$
$\frac{\sqrt{2}}{2}(e^{1/2} + e^{1/2}) = \sqrt{2}e^{1/2}$
$\sqrt{2}e^{1/2}$ (correct answer)
$2\sqrt{2}e^{1/2}$

Explanation: Using FTC:

G'(x) = e^{(\sin x)^2} \cdot \cos x - e^{(\cos x)^2} \cdot (-\sin x) = e^{\sin^2 x} \cos x + e^{\cos^2 x} \sin x

. At

x = \frac{\pi}{4}

\sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}

, so

\sin^2\frac{\pi}{4} = \cos^2\frac{\pi}{4} = \frac{1}{2}

. Therefore:

G'\left(\frac{\pi}{4}\right) = e^{1/2} \cdot \frac{\sqrt{2}}{2} + e^{1/2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}e^{1/2} + \frac{\sqrt{2}}{2}e^{1/2} = \sqrt{2}e^{1/2}

. Choice A incorrectly subtracts the terms. Choice B shows the intermediate step before combining terms. Choice D incorrectly doubles the result.

Question 11

Let $G(x) = \int_0^x (\int_1^{\cos t} u^2 du) dt$ . What is the value of $G''(\pi/2)$ ?

$-1$
$0$ (correct answer)
$1$
$\pi/2$

Explanation: This problem involves nested applications of the Fundamental Theorem of Calculus, Part 1. Let

H(t) = \int_1^{\cos t} u^2 du

. Then

G(x) = \int_0^x H(t) dt

. The first derivative is

G'(x) = H(x) = \int_1^{\cos x} u^2 du

. To find the second derivative, we differentiate

G'(x)

with respect to

x

, again using FTC1 with the chain rule:

G''(x) = \frac{d}{dx} \int_1^{\cos x} u^2 du = ((\cos x)^2) \cdot (-\sin x) = -\cos^2(x) \sin(x)

. Now, we evaluate at

x = \pi/2

G''(\pi/2) = -\cos^2(\pi/2) \sin(\pi/2) = -(0)^2(1) = 0

Question 12

Let $F(x) = \int_0^x f(t) dt$ and $G(x) = \int_5^x f(t) dt$ . If $F(5)=10$ , what is the value of $G(0)$ ?

$10$
$5$
$-5$
$-10$ (correct answer)

Explanation: We can relate

F(x)

and

G(x)

using the properties of definite integrals.

F(x) = \int_0^x f(t) dt = \int_0^5 f(t) dt + \int_5^x f(t) dt

. We recognize that

\int_0^5 f(t) dt = F(5)

and

\int_5^x f(t) dt = G(x)

. So,

F(x) = F(5) + G(x)

. We are given

F(5)=10

, so

F(x) = 10 + G(x)

. We need to find

G(0)

. Let's evaluate the relationship at

x=0

F(0) = 10 + G(0)

. By definition,

F(0) = \int_0^0 f(t) dt = 0

. So,

0 = 10 + G(0)

, which gives

G(0) = -10

. Alternatively,

G(0) = \int_5^0 f(t) dt = -\int_0^5 f(t) dt = -F(5) = -10

Question 13

The average value of the function $g(x) = e^x$ on the interval $[0, k]$ is $e-1$ . What is the value of $k$ ?

$1$ (correct answer)
$\ln(e-1)$
$e$
$\ln(e^2-e+1)$

Explanation: The formula for the average value of a function

g(x)

on an interval

[a, b]

\frac{1}{b-a} \int_a^b g(x) dx

. We are given that the average value of

g(x) = e^x

[0, k]

e-1

. Setting up the equation:

\frac{1}{k-0} \int_0^k e^x dx = e-1

. The integral of

e^x

e^x

, so we have

\frac{1}{k} [e^x]_0^k = e-1

. Evaluating the antiderivative:

\frac{1}{k} (e^k - e^0) = e-1

, which simplifies to

\frac{e^k - 1}{k} = e-1

. Therefore,

e^k - 1 = k(e-1)

. Testing

k=1

e^1 - 1 = 1(e-1)

, which gives

e-1 = e-1

. This is true, so

k=1

Question 14

Let $F(x) = \int_{x^2}^{x^3} e^{t^2} dt$ . What is the value of $F'(1)$ ?

$e$ (correct answer)
$0$
$3e - 2e$
$e^9 - e^4$

Explanation: This question requires applying the Fundamental Theorem of Calculus, Part 1, with the chain rule. The formula for the derivative of an integral with variable limits is

\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x))b'(x) - f(a(x))a'(x)

. Here,

f(t) = e^{t^2}

b(x) = x^3

, and

a(x) = x^2

. Thus,

b'(x) = 3x^2

and

a'(x) = 2x

. Applying the formula, we get

F'(x) = e^{(x^3)^2} \cdot (3x^2) - e^{(x^2)^2} \cdot (2x) = 3x^2 e^{x^6} - 2x e^{x^4}

. Now, we evaluate at

x=1

F'(1) = 3(1)^2 e^{1^6} - 2(1) e^{1^4} = 3e^1 - 2e^1 = e

Question 15

Let $f(x)$ be an invertible and differentiable function with $f(0)=1$ and $f(2)=5$ . If $\int_0^2 f(x) dx = 6$ , what is the value of $\int_1^5 f^{-1}(y) dy$ ?

$1$
$4$ (correct answer)
$9$
$10$

Explanation: This problem uses the relationship between the integral of a function and its inverse. The formula is

\int_a^b f(x) dx + \int_{f(a)}^{f(b)} f^{-1}(y) dy = b f(b) - a f(a)

. In this case,

a=0

and

b=2

. We are given

f(0)=1

and

f(2)=5

, and

\int_0^2 f(x) dx = 6

. Plugging these values into the formula:

6 + \int_1^5 f^{-1}(y) dy = 2 \cdot f(2) - 0 \cdot f(0)

. This simplifies to

6 + \int_1^5 f^{-1}(y) dy = 2 \cdot 5 - 0 = 10

. Solving for the integral, we find

\int_1^5 f^{-1}(y) dy = 10 - 6 = 4

Question 16

What is the value of the definite integral $\int_{-3}^3 |x^2 - 4| dx$ ?

$\frac{46}{3}$ (correct answer)
$\frac{32}{3}$
$6$
$0$

Explanation: The integrand

|x^2 - 4|

changes its definition based on the sign of

x^2 - 4

. The roots of

x^2 - 4 = 0

are

x = -2

and

x = 2

. For

x \in [-3, -2] \cup [2, 3]

x^2 - 4 \ge 0

. For

x \in (-2, 2)

x^2 - 4 < 0

. We must split the integral:

\int_{-3}^{-2} (x^2 - 4) dx + \int_{-2}^2 -(x^2 - 4) dx + \int_2^3 (x^2 - 4) dx

. Evaluating each part:

[\frac{x^3}{3} - 4x]_{-3}^{-2} = (-\frac{8}{3}+8) - (-9+12) = \frac{16}{3} - 3 = \frac{7}{3}

[4x - \frac{x^3}{3}]_{-2}^2 = (8-\frac{8}{3}) - (-8+\frac{8}{3}) = 2(8-\frac{8}{3}) = \frac{32}{3}

[\frac{x^3}{3} - 4x]_2^3 = (9-12) - (\frac{8}{3}-8) = -3 - (-\frac{16}{3}) = \frac{7}{3}

. The total is

\frac{7}{3} + \frac{32}{3} + \frac{7}{3} = \frac{46}{3}

Question 17

The equation of the tangent line to the curve $y = \int_x^4 \frac{dt}{1+\sqrt{t}}$ at $x=4$ is:

$y = -\frac{1}{3}(x-4)$ (correct answer)
$y = \frac{1}{3}(x-4)$
$y - 4 = -\frac{1}{3}x$
$y = 3(x-4)$

Explanation: To find the equation of a tangent line, we need a point and a slope. The point is at

x=4

. The y-coordinate is

y(4) = \int_4^4 \frac{dt}{1+\sqrt{t}} = 0

. So the point is

(4, 0)

. The slope is the derivative

y'

evaluated at

x=4

. First, rewrite the integral as

y = -\int_4^x \frac{dt}{1+\sqrt{t}}

. Using the Fundamental Theorem of Calculus, Part 1,

y' = -\frac{1}{1+\sqrt{x}}

. Now, evaluate the slope at

x=4

y'(4) = -\frac{1}{1+\sqrt{4}} = -\frac{1}{1+2} = -\frac{1}{3}

. Using the point-slope form

y - y_1 = m(x - x_1)

, we get

y - 0 = -\frac{1}{3}(x - 4)

, which is

y = -\frac{1}{3}(x-4)

Question 18

Let $f$ be a continuous function. If $\int_0^3 f(x) dx = 6$ and the average value of $f$ on $[3, 8]$ is 4, what is the average value of $f$ on $[0, 8]$ ?

$5$
$10$
$4.5$
$3.25$ (correct answer)

Explanation: When you encounter problems involving average values over different intervals, you need to connect the definition of average value with properties of definite integrals. The average value of a function

f

on interval

[a,b]

\frac{1}{b-a}\int_a^b f(x)dx

. You're given that

\int_0^3 f(x)dx = 6

and the average value on

[3,8]

is 4. Since the average value on

[3,8]

is 4, you can find:

\frac{1}{8-3}\int_3^8 f(x)dx = 4

, so

\int_3^8 f(x)dx = 4 \times 5 = 20

. Now you can find the average value on

[0,8]

. Using the additive property of integrals:

\int_0^8 f(x)dx = \int_0^3 f(x)dx + \int_3^8 f(x)dx = 6 + 20 = 26

. Therefore, the average value on

[0,8]

\frac{26}{8-0} = \frac{26}{8} = 3.25

, which is answer D. Looking at the wrong answers: A (5) might come from averaging the two given average values, but this ignores that the intervals have different lengths. B (10) could result from incorrectly calculating

\frac{20}{2}

. C (4.5) might come from adding the average values and dividing by 2, again not accounting for interval lengths. Study tip: Remember that average values over combined intervals depend on both the individual averages AND the lengths of each interval. You can't simply average the average values unless the intervals are equal in length.

Question 19

What is the value of $\lim_{x \to 0} \frac{1}{x} \int_0^x \cos(t^2 + t) dt$ ?

$0$
$1$ (correct answer)
The limit does not exist.
$\sin(x^2+x)$

Explanation: This limit is in the form of the definition of a derivative. Let

G(x) = \int_0^x \cos(t^2 + t) dt

. The expression can be rewritten as

\lim_{x \to 0} \frac{G(x) - G(0)}{x - 0}

, since

G(0) = \int_0^0 \cos(t^2 + t) dt = 0

. This is the definition of

G'(0)

. By the Fundamental Theorem of Calculus, Part 1,

G'(x) = \cos(x^2 + x)

. Therefore, the value of the limit is

G'(0) = \cos(0^2 + 0) = \cos(0) = 1

Question 20

Let $f(x)$ be a continuous function. If $\int_0^4 f(x) dx = 10$ , what is the value of $\int_0^2 f(2x) dx$ ?

$5$ (correct answer)
$10$
$20$
Cannot be determined from the given information.

Explanation: This problem requires using u-substitution on a definite integral. Let's evaluate

\int_0^2 f(2x) dx

. Let

u = 2x

. Then

du = 2 dx

, or

dx = \frac{1}{2} du

. We also need to change the limits of integration. When

x=0

u = 2(0) = 0

. When

x=2

u = 2(2) = 4

. Substituting into the integral, we get

\int_0^4 f(u) \cdot \frac{1}{2} du = \frac{1}{2} \int_0^4 f(u) du

. Since we are given

\int_0^4 f(x) dx = 10

, the value of

\int_0^4 f(u) du

is also 10 (the variable of integration does not change the value of the definite integral). Therefore, the result is

\frac{1}{2} (10) = 5