Properties of Definite Integrals - Calculus 1

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 $\int_{3}^{1} f(x) \,dx$ ?
Answer: $-5$ 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, $\int_{3}^{1} f(x) \,dx = -\int_{1}^{3} f(x) \,dx = -5$ .
Given that $\int_{a}^{b} f(x) \,dx = M$ and $\int_{a}^{b} g(x) \,dx = N$ . Which of the following is equal to $\int_{b}^{a} [2f(x) - g(x)] \,dx$ ?
Answer: $N - 2M$ 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 $N - 2M$ .
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?
Answer: $\int_{0}^{5} |f(x)| \,dx \ge 4$ 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.
Given $\int_{2}^{8} f(x) \,dx = 12$ and $\int_{5}^{2} f(x) \,dx = 3$ . What is the value of $\int_{5}^{8} f(x) \,dx$ ?
Answer: $15$ 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 $\int_{5}^{2}$ to $\int_{2}^{5}$ . 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.
Let $f(x)$ be a continuous function. If $\int_{0}^{9} f(x) \,dx = 12$ , what is the value of $\int_{0}^{3} x f(x^2) \,dx$ ?
Answer: $6$ 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.
Suppose $\int_{0}^{4} f(x) \,dx = 10$ and $\int_{0}^{4} g(x) \,dx = -2$ . What is the value of $\int_{0}^{4} [2f(x) - 3g(x) + 1] \,dx$ ?
Answer: $30$ 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 $20 + 6 + 4 = 30$ .
Given that $f(x)$ is a continuous function and $f(x) \ge 3$ for $x \in [1, 5]$ , what is the smallest possible value of $\int_{1}^{5} [f(x) - 1] \,dx$ ?
Answer: $8$ If $f(x) \ge 3$ , then the integrand $g(x) = f(x) - 1$ satisfies $g(x) \ge 3 - 1 = 2$ . Using the comparison property for integrals, the smallest possible value for $\int_{1}^{5} g(x) \,dx$ occurs when $g(x)$ is at its minimum value throughout the interval. The lower bound is $m(b-a)$ , where $m=2$ , $a=1$ , and $b=5$ . So, the minimum value is $2(5-1) = 2(4) = 8$ .
If $\int_{1}^{3} f(x) \,dx = 8$ , what is the value of $\int_{0}^{2} f(x+1) \,dx$ ?
Answer: $8$ 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.
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 $\int_{-1}^{2} f(x) \,dx$ ?
Answer: $2$ 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.
If $\int_{0}^{k} (4x - 5) \,dx = 0$ for some constant $k$ , what is a possible non-zero value of $k$ ?
Answer: $5/2$ 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.
A function $f(x)$ is defined as $f(x) = 3$ for $0 \le x < 2$ and $f(x) = -1$ for $2 \le x \le 5$ . What is the value of $\int_{0}^{5} f(x) \,dx$ ?
Answer: $3$ We must split the integral at the point $x=2$ where the function definition changes: $\int_{0}^{5} f(x) \,dx = \int_{0}^{2} f(x) \,dx + \int_{2}^{5} f(x) \,dx$ . Substitute the corresponding function values: $\int_{0}^{2} 3 \,dx + \int_{2}^{5} -1 \,dx$ . Evaluate each integral of a constant: $3(2-0) + (-1)(5-2) = 3(2) - 1(3) = 6 - 3 = 3$ .
Given $\int_{1}^{2} f(x)dx = 3$ and $\int_{2}^{4} f(x)dx = 5$ . What is the value of $\int_{4}^{1} (2f(x)+3)dx$ ?
Answer: $-25$ 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 $2(-8) + (-9) = -16 - 9 = -25$ .
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 $\int_{1}^{3} f(x) \,dx$ ?
Answer: $5$ 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 $I_f = 5$ .
The average value of a continuous function $g(x)$ on the interval $[-2, 4]$ is 5. What is the value of $\int_{-2}^{4} g(x) \,dx$ ?
Answer: $30$ 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.
If $f(x)$ is a continuous and non-positive function on $[a, b]$ with $a<b$ , which of the following must be true?
Answer: $\int_{a}^{b} f(x) \,dx \le 0$ Non-positive means $f(x) \le 0$ for all $x$ in $[a,b]$ . 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 $-\int_a^b f(x) \,dx \ge 0$ .
Suppose $\int_{0}^{4} f(x) \,dx = 6$ . What is the value of $\int_{0}^{2} f(2x) \,dx$ ?
Answer: $3$ 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 $\frac{1}{2}\int_{0}^{4} f(x) \,dx = \frac{1}{2}(6) = 3$ .
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 $I = \int_{1}^{4} f(x) \,dx$ ?
Answer: $6 \le I \le 15$ 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 $6 \le I \le 15$ .
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 $\int_{2}^{8} f(x) \,dx$ ?
Answer: $8$ 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 $-3 + 7 + 4 = 8$ .
Let $f(x)$ be a continuous function. If $\int_{-2}^{2} f(x) \,dx = 10$ , what is the value of $\int_{0}^{4} f(x-2) \,dx$ ?
Answer: $10$ 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.
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 $k$ ?
Answer: $5$ 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.

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