Calculus 1 Quiz: Properties Of Definite Integrals

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Properties Of Definite IntegralsQuestion 1 of 20

Suppose $f(x)$ is a continuous function such that $2 \le f(x) \le 5$ for all $x$ in the interval $[1, 4]$ . Which of the following statements must be true about $I = \int_{1}^{4} f(x) \,dx$ ?

2 \le I \le 5

I = 10.5

3 \le I \le 12

6 \le I \le 15

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Calculus 1 Quiz

Calculus 1 Quiz: Properties Of Definite Integrals

Practice Properties Of Definite Integrals in Calculus 1 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 Properties Of Definite Integrals, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

Suppose $f(x)$ is a continuous function such that $2 \le f(x) \le 5$ for all $x$ in the interval $[1, 4]$ . Which of the following statements must be true about $I = \int_{1}^{4} f(x) \,dx$ ?

$2 \le I \le 5$
$I = 10.5$
$3 \le I \le 12$
$6 \le I \le 15$ (correct answer)

Explanation: According to the comparison property of definite integrals, if

m \le f(x) \le M

for

a \le x \le b

, then

m(b-a) \le \int_{a}^{b} f(x) \,dx \le M(b-a)

. In this case,

m=2

M=5

a=1

, and

b=4

. The length of the interval is

b-a = 4-1 = 3

. Therefore, the integral

I

must satisfy

2(3) \le I \le 5(3)

, which simplifies to

6 \le I \le 15

Question 2

Let $f(x)$ be a continuous function. If $\int_{-2}^{2} f(x) \,dx = 10$ , what is the value of $\int_{0}^{4} f(x-2) \,dx$ ?

$5$
$8$
$10$ (correct answer)
$12$

Explanation: This problem can be solved using a u-substitution, which is a property related to transformations of integrals. Let

u = x-2

. Then

du = dx

. We must also change the limits of integration. When

x=0

u = 0-2 = -2

. When

x=4

u = 4-2 = 2

. Substituting these into the integral gives

\int_{-2}^{2} f(u) \,du

. Since the variable of integration is a dummy variable, this is the same as

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

, which is given to be 10.

Question 3

If $\int_{0}^{k} (4x - 5) \,dx = 0$ for some constant $k$ , what is a possible non-zero value of $k$ ?

$5/4$
$4/5$
$5/2$ (correct answer)
$2/5$

Explanation: First, evaluate the definite integral:

\int_{0}^{k} (4x - 5) \,dx = [2x^2 - 5x]_{0}^{k}

. Evaluating at the limits gives

(2k^2 - 5k) - (2(0)^2 - 5(0)) = 2k^2 - 5k

. Set this expression equal to zero:

2k^2 - 5k = 0

. Factor out

k

to get

k(2k - 5) = 0

. The solutions are

k=0

and

2k-5=0

, which gives

k=5/2

. The question asks for the non-zero value.

Question 4

Given that $f(x)$ is a continuous function and $\int_{0}^{5} f(x) \,dx = -4$ . Which of the following statements about $\int_{0}^{5} |f(x)| \,dx$ must be true?

$\int_{0}^{5} |f(x)| \,dx = 4$
$\int_{0}^{5} |f(x)| \,dx = -4$
$\int_{0}^{5} |f(x)| \,dx \le 4$
$\int_{0}^{5} |f(x)| \,dx \ge 4$ (correct answer)

Explanation: When you encounter integrals involving absolute value functions, think about how absolute values affect the relationship between

\int f(x) \,dx

and

\int |f(x)| \,dx

. The key insight is that taking absolute values can only increase (or maintain) the total "area under the curve." Since

\int_{0}^{5} f(x) \,dx = -4

, we know that the net signed area is negative, meaning the total area below the x-axis exceeds the area above it. When we integrate

|f(x)|

, we're finding the total area regardless of sign—all negative contributions become positive. Here's why answer D is correct: The integral

\int_{0}^{5} |f(x)| \,dx

must be at least 4. To see why, imagine the most extreme case where

f(x) \leq 0

throughout the entire interval. Then

|f(x)| = -f(x)

, so

\int_{0}^{5} |f(x)| \,dx = -\int_{0}^{5} f(x) \,dx = -(-4) = 4

. If

f(x)

has any positive values, the integral of

|f(x)|

would be even larger. Answer A is wrong because equality only holds in the special case where

f(x) \leq 0

everywhere. Answer B is impossible since absolute values are non-negative, making their integral non-negative. Answer C suggests the integral could be less than 4, but we've shown it must be at least 4. Study tip: Remember that

\int |f(x)| \,dx \geq \left|\int f(x) \,dx\right|

always holds. When the original integral is negative, the absolute value integral is at least the magnitude of that negative value.

Question 5

Given $\int_{0}^{2} f(x) \,dx = 3$ , $\int_{5}^{0} f(x) \,dx = -7$ , and $\int_{5}^{8} f(x) \,dx = 4$ . What is the value of $\int_{2}^{8} f(x) \,dx$ ?

$-6$
$0$
$8$ (correct answer)
$14$

Explanation: First, use the property of reversing limits:

\int_{5}^{0} f(x) \,dx = -7

implies

\int_{0}^{5} f(x) \,dx = 7

. We want to find

\int_{2}^{8} f(x) \,dx

. We can express this integral using the additivity property of intervals:

\int_{2}^{8} f(x) \,dx = \int_{2}^{0} f(x) \,dx + \int_{0}^{5} f(x) \,dx + \int_{5}^{8} f(x) \,dx

. We know

\int_{2}^{0} f(x) \,dx = -\int_{0}^{2} f(x) \,dx = -3

. Substituting all the values, we get

-3 + 7 + 4 = 8

Question 6

Given $\int_{2}^{8} f(x) \,dx = 12$ and $\int_{5}^{2} f(x) \,dx = 3$ . What is the value of $\int_{5}^{8} f(x) \,dx$ ?

$-9$
$9$
$-15$
$15$ (correct answer)

Explanation: When you encounter definite integrals with different limits of integration, you need to use the additive property of integrals and understand how reversing limits affects the sign. First, notice that

\int_{5}^{2} f(x) \,dx = 3

. When you reverse the limits of integration, the integral changes sign:

\int_{2}^{5} f(x) \,dx = -\int_{5}^{2} f(x) \,dx = -3

. Now you can use the additive property:

\int_{2}^{8} f(x) \,dx = \int_{2}^{5} f(x) \,dx + \int_{5}^{8} f(x) \,dx

. Substituting the known values:

12 = (-3) + \int_{5}^{8} f(x) \,dx

. Solving for the unknown integral:

\int_{5}^{8} f(x) \,dx = 12 - (-3) = 12 + 3 = 15

. Choice A gives

-9

, which you'd get if you incorrectly calculated

12 - 3 - 3 = 6

and then made an additional sign error. Choice B gives 9, which results from the error

12 - 3 = 9

—forgetting to reverse the sign when flipping the limits from

\int_{5}^{2}

\int_{2}^{5}

. Choice C gives

-15

, which comes from incorrectly adding the integrals as

12 + 3 = 15

but then applying a wrong negative sign. The correct answer is D: 15. Strategy tip: Always remember that

\int_{a}^{b} f(x) \,dx = -\int_{b}^{a} f(x) \,dx

, and use the additive property to break complex limits into manageable pieces. Draw a number line to visualize the intervals if needed.

Question 7

Given that $\int_{a}^{b} f(x) \,dx = M$ and $\int_{a}^{b} g(x) \,dx = N$ . Which of the following is equal to $\int_{b}^{a} [2f(x) - g(x)] \,dx$ ?

$2M - N$
$N - 2M$ (correct answer)
$M - 2N$
$2N - M$

Explanation: Start by reversing the limits of integration, which introduces a negative sign:

\int_{b}^{a} [2f(x) - g(x)] \,dx = -\int_{a}^{b} [2f(x) - g(x)] \,dx

. Next, apply the linearity property:

-\left[ 2\int_{a}^{b} f(x) \,dx - \int_{a}^{b} g(x) \,dx \right]

. Distribute the negative sign:

-2\int_{a}^{b} f(x) \,dx + \int_{a}^{b} g(x) \,dx

. Finally, substitute the given values M and N:

-2M + N

, which is equivalent to

N - 2M

Question 8

If $\int_{2}^{5} f(x) \,dx = -4$ and $\int_{2}^{5} (f(x) + k) \,dx = 11$ , where $k$ is a constant, what is the value of $k$ ?

$5$ (correct answer)
$3$
$7$
$15$

Explanation: When you encounter definite integrals with added constants, remember that integration distributes over addition, and constants can be factored out and integrated separately. Let's work with the given information systematically. You know that

\int_{2}^{5} f(x) \,dx = -4

and

\int_{2}^{5} (f(x) + k) \,dx = 11

. Using the linearity property of integrals, you can split the second integral:

\int_{2}^{5} (f(x) + k) \,dx = \int_{2}^{5} f(x) \,dx + \int_{2}^{5} k \,dx

Since

k

is a constant,

\int_{2}^{5} k \,dx = k \cdot (5-2) = 3k

Substituting the known values:

11 = -4 + 3k

Solving for

k

15 = 3k

, so

k = 5

Therefore, A)

5

is correct. Let's examine why the other answers are wrong. B)

3

would give us

\int_{2}^{5} (f(x) + k) \,dx = -4 + 3(3) = 5

, not 11. C)

7

would yield

-4 + 3(7) = 17

, which exceeds our target. D)

15

would produce

-4 + 3(15) = 41

, far too large. Key strategy: When you see integrals with added constants, immediately think about splitting them using linearity properties. Remember that integrating a constant over an interval gives you the constant times the length of the interval. This type of problem tests your understanding of integral properties more than complex calculations.

Question 9

Let $f(x)$ be an even function such that $\int_{0}^{2} f(x) \,dx = 5$ . What is the value of $\int_{-2}^{2} (3f(x) - 2) \,dx$ ?

$11$
$13$
$22$ (correct answer)
$26$

Explanation: First, use the linearity property:

\int_{-2}^{2} (3f(x) - 2) \,dx = 3\int_{-2}^{2} f(x) \,dx - \int_{-2}^{2} 2 \,dx

. Since

f(x)

is an even function,

\int_{-2}^{2} f(x) \,dx = 2\int_{0}^{2} f(x) \,dx = 2(5) = 10

. The integral of the constant is

\int_{-2}^{2} 2 \,dx = 2(2 - (-2)) = 2(4) = 8

. Combining these results, we get

3(10) - 8 = 30 - 8 = 22

Question 10

Suppose $\int_{0}^{4} f(x) \,dx = 6$ . What is the value of $\int_{0}^{2} f(2x) \,dx$ ?

$3$ (correct answer)
$4$
$6$
$12$

Explanation: Use u-substitution. Let

u = 2x

. Then

du = 2dx

, which means

dx = \frac{1}{2}du

. Change the limits of integration: when

x=0

u=2(0)=0

; when

x=2

u=2(2)=4

. The integral becomes

\int_{0}^{4} f(u) \frac{1}{2}du

. Using the constant multiple property, this is

\frac{1}{2}\int_{0}^{4} f(u) \,du

. Since

u

is a dummy variable, this is equal to

\frac{1}{2}\int_{0}^{4} f(x) \,dx = \frac{1}{2}(6) = 3

Question 11

Let $f(x)$ be an odd function and $g(x)$ be an even function. If $\int_{0}^{3} f(x) \,dx = 4$ and $\int_{0}^{3} g(x) \,dx = 5$ , what is the value of $\int_{-3}^{3} [f(x) + g(x)] \,dx$ ?

$0$
$9$
$10$ (correct answer)
$18$

Explanation: First, split the integral:

\int_{-3}^{3} [f(x) + g(x)] \,dx = \int_{-3}^{3} f(x) \,dx + \int_{-3}^{3} g(x) \,dx

. For an odd function

f

, the integral over a symmetric interval is zero, so

\int_{-3}^{3} f(x) \,dx = 0

. For an even function

g

, the integral over a symmetric interval is twice the integral over half the interval, so

\int_{-3}^{3} g(x) \,dx = 2\int_{0}^{3} g(x) \,dx = 2(5) = 10

. Therefore, the total value is

0 + 10 = 10

Question 12

Suppose $\int_{1}^{3} [f(x) + g(x)] \,dx = 6$ and $\int_{1}^{3} [2f(x) - g(x)] \,dx = 9$ . What is the value of $\int_{1}^{3} f(x) \,dx$ ?

$1$
$3$
$5$ (correct answer)
$7.5$

Explanation: Let

I_f = \int_{1}^{3} f(x) \,dx

and

I_g = \int_{1}^{3} g(x) \,dx

. Using the linearity property, the given information can be written as a system of two linear equations: (1)

I_f + I_g = 6

and (2)

2I_f - I_g = 9

. Adding the two equations together yields

(I_f + I_g) + (2I_f - I_g) = 6 + 9

, which simplifies to

3I_f = 15

. Solving for

I_f

gives

I_f = 5

Question 13

A continuous function $f(x)$ satisfies $-x^2 \le f(x) \le x^2$ for all $x$ . Which of the following is a possible value for $\int_{-1}^{2} f(x) \,dx$ ?

$-4$
$2$ (correct answer)
$3.5$
$4$

Explanation: By the comparison property of integrals,

\int_{-1}^{2} -x^2 \,dx \le \int_{-1}^{2} f(x) \,dx \le \int_{-1}^{2} x^2 \,dx

. We calculate the bounds.

\int x^2 \,dx = \frac{x^3}{3}

. So,

\int_{-1}^{2} x^2 \,dx = [\frac{x^3}{3}]_{-1}^{2} = \frac{2^3}{3} - \frac{(-1)^3}{3} = \frac{8}{3} - (-\frac{1}{3}) = \frac{9}{3} = 3

. The lower bound is

\int_{-1}^{2} -x^2 \,dx = -3

. Therefore, the value of

\int_{-1}^{2} f(x) \,dx

must be in the interval

[-3, 3]

. Of the choices provided, only 2 lies within this range.

Question 14

If $f(x)$ is a continuous and non-positive function on $[a, b]$ with $a<b$ , which of the following must be true?

$\int_{a}^{b} f(x) \,dx > 0$
$\int_{a}^{b} f(x) \,dx = 0$
$\int_{a}^{b} f(x) \,dx \le 0$ (correct answer)
$\int_{b}^{a} f(x) \,dx \le 0$

Explanation: Non-positive means

f(x) \le 0

for all

x

[a,b]

. By the comparison property of integrals, if

f(x) \le g(x)

on an interval, then

\int_a^b f(x) \,dx \le \int_a^b g(x) \,dx

. Let

g(x) = 0

. Then

\int_a^b f(x) \,dx \le \int_a^b 0 \,dx = 0

. Therefore,

\int_a^b f(x) \,dx \le 0

. Choice D is incorrect because

\int_b^a f(x) \,dx = -\int_a^b f(x) \,dx

. Since

\int_a^b f(x) \,dx \le 0

, then

-\int_a^b f(x) \,dx \ge 0

Question 15

Suppose $\int_{0}^{4} f(x) \,dx = 10$ and $\int_{0}^{4} g(x) \,dx = -2$ . What is the value of $\int_{0}^{4} [2f(x) - 3g(x) + 1] \,dx$ ?

$14$
$26$
$27$
$30$ (correct answer)

Explanation: Using the linearity property of definite integrals:

\int_{0}^{4} [2f(x) - 3g(x) + 1] \,dx = 2\int_{0}^{4} f(x) \,dx - 3\int_{0}^{4} g(x) \,dx + \int_{0}^{4} 1 \,dx

. Substitute the given values:

2(10) - 3(-2) + \int_{0}^{4} 1 \,dx

. The last term is the integral of a constant, which is

1(4-0) = 4

. The final calculation is

20 + 6 + 4 = 30

Question 16

If $\int_{1}^{3} f(x) \,dx = 8$ , what is the value of $\int_{0}^{2} f(x+1) \,dx$ ?

$7$
$8$ (correct answer)
$9$
$16$

Explanation: We use u-substitution. Let

u = x+1

. Then

du = dx

. We also change the limits of integration: when

x=0

u=0+1=1

; when

x=2

u=2+1=3

. The integral becomes

\int_{1}^{3} f(u) \,du

. Since the variable of integration is a dummy variable, this is equivalent to

\int_{1}^{3} f(x) \,dx

, which is given as 8.

Question 17

Given $\int_{1}^{2} f(x)dx = 3$ and $\int_{2}^{4} f(x)dx = 5$ . What is the value of $\int_{4}^{1} (2f(x)+3)dx$ ?

$-25$ (correct answer)
$-13$
$19$
$25$

Explanation: First, find

\int_{1}^{4} f(x)dx

using interval additivity:

\int_{1}^{4} f(x)dx = \int_{1}^{2} f(x)dx + \int_{2}^{4} f(x)dx = 3 + 5 = 8

. Next, evaluate the required integral. By linearity,

\int_{4}^{1} (2f(x)+3)dx = 2\int_{4}^{1} f(x)dx + \int_{4}^{1} 3dx

. Reversing the limits,

\int_{4}^{1} f(x)dx = -\int_{1}^{4} f(x)dx = -8

. The integral of the constant is

\int_{4}^{1} 3dx = 3(1-4) = 3(-3) = -9

. Combining these gives

2(-8) + (-9) = -16 - 9 = -25

Question 18

Let f be a continuous function. If $\int_{1}^{5} f(x) \,dx = 8$ and $\int_{3}^{5} f(x) \,dx = 3$ , what is the value of $\int_{3}^{1} f(x) \,dx$ ?

$-11$
$-5$ (correct answer)
$5$
$11$

Explanation: Using the interval additivity property,

\int_{1}^{5} f(x) \,dx = \int_{1}^{3} f(x) \,dx + \int_{3}^{5} f(x) \,dx

. Substituting the given values, we get

8 = \int_{1}^{3} f(x) \,dx + 3

, which implies

\int_{1}^{3} f(x) \,dx = 5

. The question asks for

\int_{3}^{1} f(x) \,dx

. Using the property of reversing the limits of integration,

\int_{3}^{1} f(x) \,dx = -\int_{1}^{3} f(x) \,dx = -5

Question 19

Let $f(x)$ be a continuous function. If $\int_{0}^{9} f(x) \,dx = 12$ , what is the value of $\int_{0}^{3} x f(x^2) \,dx$ ?

$6$ (correct answer)
$4$
$12$
$24$

Explanation: When you encounter an integral with a composite function like

f(x^2)

, think substitution. The key insight is recognizing how the given integral relates to what you're asked to find through a change of variables. To evaluate

\int_{0}^{3} x f(x^2) \,dx

, use the substitution

u = x^2

. Then

du = 2x \,dx

, which means

x \,dx = \frac{1}{2} du

. When

x = 0,u = 0

. When

x = 3,u = 9

. Substituting gives you:

\int_{0}^{3} x f(x^2) \,dx = \int_{0}^{9} f(u) \cdot \frac{1}{2} \,du = \frac{1}{2} \int_{0}^{9} f(u) \,du

Since the variable of integration is just a dummy variable,

\int_{0}^{9} f(u) \,du = \int_{0}^{9} f(x) \,dx = 12

. Therefore:

\int_{0}^{3} x f(x^2) \,dx = \frac{1}{2} \cdot 12 = 6

The answer is (A) 6. (B) 4 likely comes from incorrectly thinking the factor should be

\frac{1}{3}

instead of

\frac{1}{2}

. (C) 12 is the trap of assuming the integrals are equal without accounting for the substitution's Jacobian factor. (D) 24 results from mistakenly using a factor of 2 instead of

\frac{1}{2}

, perhaps from confusing

du = 2x \,dx

with the conversion factor. Remember: in u-substitution problems, always carefully track how

dx

transforms, including any multiplicative factors that arise from the derivative of your substitution.

Question 20

The average value of a continuous function $g(x)$ on the interval $[-2, 4]$ is 5. What is the value of $\int_{-2}^{4} g(x) \,dx$ ?

$5/6$
$10$
Cannot be determined from the given information.
$30$ (correct answer)

Explanation: This question tests your understanding of the relationship between average value and definite integrals. When you see "average value" paired with an integral, think about how these concepts connect through the Mean Value Theorem for Integrals. The average value of a continuous function

g(x)

on interval

[a,b]

is defined as:

\text{Average value} = \frac{1}{b-a} \int_a^b g(x) \, dx

Given that the average value is 5 on interval

[-2, 4]

, you can substitute into this formula:

5 = \frac{1}{4-(-2)} \int_{-2}^4 g(x) \, dx = \frac{1}{6} \int_{-2}^4 g(x) \, dx

Solving for the integral:

\int_{-2}^4 g(x) \, dx = 5 \times 6 = 30

Looking at the wrong answers: Choice A (

5/6

) represents the reciprocal error—dividing 5 by 6 instead of multiplying. This happens when students confuse the formula direction. Choice B (

10

) might come from incorrectly calculating the interval length as 2 instead of 6, then multiplying

5 \times 2

. Choice C suggests the integral can't be determined, but the average value formula directly connects these quantities—if you know the average value and interval length, you can always find the integral. Remember this key relationship:

\int_a^b f(x) \, dx = (\text{average value}) \times (\text{interval length})

. This formula is your bridge between average value problems and definite integrals, and it works in both directions depending on what you're asked to find.