IB Mathematics: Analysis and Approaches Quiz: Integration Techniques
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Integration TechniquesQuestion 1 of 20

Find ∫cos⁡(4x−π2)dx\int \cos(4x - \frac{\pi}{2}) dx.

−14sin⁡(4x−π2)+C-\frac{1}{4}\sin(4x - \frac{\pi}{2}) + C
−4sin⁡(4x−π2)+C-4\sin(4x - \frac{\pi}{2}) + C
14sin⁡(4x−π2)+C\frac{1}{4}\sin(4x - \frac{\pi}{2}) + C
sin⁡(4x−π2)+C\sin(4x - \frac{\pi}{2}) + C
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Integration Techniques

Practice Integration Techniques in IB Mathematics: Analysis and Approaches 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 Integration Techniques, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Question 1

Find ∫cos⁡(4x−π2)dx\int \cos(4x - \frac{\pi}{2}) dx.

  1. −14sin⁡(4x−π2)+C-\frac{1}{4}\sin(4x - \frac{\pi}{2}) + C
  2. −4sin⁡(4x−π2)+C-4\sin(4x - \frac{\pi}{2}) + C
  3. 14sin⁡(4x−π2)+C\frac{1}{4}\sin(4x - \frac{\pi}{2}) + C (correct answer)
  4. sin⁡(4x−π2)+C\sin(4x - \frac{\pi}{2}) + C
Explanation: This requires the reverse chain rule. The integral of cos⁡(u)\cos(u) is sin⁡(u)\sin(u). The inner function is u=4x−π2u = 4x - \frac{\pi}{2}, and its derivative is u′=4u' = 4. We must divide the result by this derivative. Thus, ∫cos⁡(4x−π2)dx=14sin⁡(4x−π2)+C\int \cos(4x - \frac{\pi}{2}) dx = \frac{1}{4}\sin(4x - \frac{\pi}{2}) + C.

Question 2

Given that ∫0k(2x−3)dx=4\int_0^k (2x - 3) dx = 4, and k>0k > 0, find the value of kk.

  1. -1
  2. 1
  3. 4 (correct answer)
  4. 7
Explanation: First, evaluate the definite integral in terms of kk. The antiderivative of 2x−32x-3 is x2−3xx^2 - 3x. Evaluate this from 0 to kk: [x2−3x]0k=(k2−3k)−(02−3(0))=k2−3k[x^2 - 3x]_0^k = (k^2 - 3k) - (0^2 - 3(0)) = k^2 - 3k. Now set this expression equal to 4: k2−3k=4k^2 - 3k = 4. Rearrange into a standard quadratic equation: k2−3k−4=0k^2 - 3k - 4 = 0. Factor the quadratic: (k−4)(k+1)=0(k-4)(k+1) = 0. The possible values for kk are k=4k=4 and k=−1k=-1. Since the problem states that k>0k > 0, the correct answer is k=4k=4.

Question 3

Find ∫(x−2)2xdx\int \frac{(x-2)^2}{\sqrt{x}} dx.

  1. 25x5/2−83x3/2+8x+C\frac{2}{5}x^{5/2} - \frac{8}{3}x^{3/2} + 8\sqrt{x} + C (correct answer)
  2. 25x5/2+8x+C\frac{2}{5}x^{5/2} + 8\sqrt{x} + C
  3. 13(x−2)32x+C\frac{\frac{1}{3}(x-2)^3}{2\sqrt{x}} + C
  4. 25x5/2−2x3/2+4x+C\frac{2}{5}x^{5/2} - 2x^{3/2} + 4\sqrt{x} + C
Explanation: First, expand the numerator and rewrite the expression with fractional exponents: (x−2)2x=x2−4x+4x1/2\frac{(x-2)^2}{\sqrt{x}} = \frac{x^2 - 4x + 4}{x^{1/2}}. Then, divide each term in the numerator by x1/2x^{1/2}: x3/2−4x1/2+4x−1/2x^{3/2} - 4x^{1/2} + 4x^{-1/2}. Now, integrate term by term using the power rule ∫xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C: ∫(x3/2−4x1/2+4x−1/2)dx=x5/25/2−4x3/23/2+4x1/21/2+C\int (x^{3/2} - 4x^{1/2} + 4x^{-1/2}) dx = \frac{x^{5/2}}{5/2} - \frac{4x^{3/2}}{3/2} + \frac{4x^{1/2}}{1/2} + C. Simplifying gives: 25x5/2−83x3/2+8x1/2+C=25x5/2−83x3/2+8x+C\frac{2}{5}x^{5/2} - \frac{8}{3}x^{3/2} + 8x^{1/2} + C = \frac{2}{5}x^{5/2} - \frac{8}{3}x^{3/2} + 8\sqrt{x} + C.

Question 4

Given that the derivative of a function is f′(x)=3x2−4x2f'(x) = 3x^2 - \frac{4}{x^2} and the function passes through the point (2,9)(2, 9), find f(x)f(x).

  1. f(x)=x3+4x−1f(x) = x^3 + \frac{4}{x} - 1 (correct answer)
  2. f(x)=x3−4x+3f(x) = x^3 - \frac{4}{x} + 3
  3. f(x)=6x+8x3−5f(x) = 6x + \frac{8}{x^3} - 5
  4. f(x)=x3+4x+19f(x) = x^3 + \frac{4}{x} + 19
Explanation: To find f(x)f(x), we must integrate f′(x)f'(x). First, rewrite f′(x)f'(x) as 3x2−4x−23x^2 - 4x^{-2}. The integral is f(x)=∫(3x2−4x−2)dx=3x33−4x−1−1+C=x3+4x−1+C=x3+4x+Cf(x) = \int (3x^2 - 4x^{-2}) dx = \frac{3x^3}{3} - \frac{4x^{-1}}{-1} + C = x^3 + 4x^{-1} + C = x^3 + \frac{4}{x} + C. Now use the given point (2,9)(2, 9) to solve for the constant of integration, CC. Substitute x=2x=2 and f(x)=9f(x)=9: 9=23+42+C  ⟹  9=8+2+C  ⟹  9=10+C  ⟹  C=−19 = 2^3 + \frac{4}{2} + C \implies 9 = 8 + 2 + C \implies 9 = 10 + C \implies C = -1. Therefore, the function is f(x)=x3+4x−1f(x) = x^3 + \frac{4}{x} - 1.

Question 5

Find ∫53x+4dx\int \frac{5}{\sqrt{3x+4}} dx.

  1. 103x+4+C10\sqrt{3x+4} + C
  2. −109(3x+4)−3/2+C-\frac{10}{9}(3x+4)^{-3/2} + C
  3. 53ln⁡∣3x+4∣+C\frac{5}{3}\ln|\sqrt{3x+4}| + C
  4. 1033x+4+C\frac{10}{3}\sqrt{3x+4} + C (correct answer)
Explanation: First, rewrite the integrand with a negative fractional exponent: ∫5(3x+4)−1/2dx\int 5(3x+4)^{-1/2} dx. Apply the reverse chain rule. The antiderivative of u−1/2u^{-1/2} is u1/21/2=2u1/2\frac{u^{1/2}}{1/2} = 2u^{1/2}. The inner function is 3x+43x+4, and its derivative is 3. We must divide by 3. So, the integral is 5⋅(3x+4)1/21/2⋅13+C=5⋅23x+4⋅13+C=1033x+4+C5 \cdot \frac{(3x+4)^{1/2}}{1/2} \cdot \frac{1}{3} + C = 5 \cdot 2\sqrt{3x+4} \cdot \frac{1}{3} + C = \frac{10}{3}\sqrt{3x+4} + C.

Question 6

The area of the region bounded by the curve y=x2−4y = x^2 - 4, the x-axis, from x=0x=0 to x=2x=2 is given by which expression?

  1. ∫02(x2−4)dx\int_0^2 (x^2-4) dx
  2. ∫02(4−x2)dx\int_0^2 (4-x^2) dx (correct answer)
  3. ∫−22(x2−4)dx\int_{-2}^2 (x^2-4) dx
  4. [x33−4x]02[\frac{x^3}{3} - 4x]_0^2
Explanation: The function y=x2−4y = x^2 - 4 is non-positive on the interval [0,2][0, 2]. For example, at x=1x=1, y=1−4=−3y = 1-4 = -3. To find the geometric area, we must integrate the absolute value of the function, ∣x2−4∣|x^2 - 4|. Since x2−4≤0x^2 - 4 \le 0 on [0,2][0, 2], this is equivalent to integrating −(x2−4)-(x^2 - 4), which is 4−x24 - x^2. Therefore, the correct expression for the area is ∫02(4−x2)dx\int_0^2 (4-x^2) dx. Option A would give a negative result, which is the signed area, not the geometric area.

Question 7

The area of the region bounded by the curve y=x2−4y = x^2 - 4, the x-axis, from x=0x=0 to x=2x=2 is given by which expression?

  1. ∫02(x2−4)dx\int_0^2 (x^2-4) dx
  2. ∫02(4−x2)dx\int_0^2 (4-x^2) dx (correct answer)
  3. ∫−22(x2−4)dx\int_{-2}^2 (x^2-4) dx
  4. [x33−4x]02[\frac{x^3}{3} - 4x]_0^2
Explanation: The function y=x2−4y = x^2 - 4 is non-positive on the interval [0,2][0, 2]. For example, at x=1x=1, y=1−4=−3y = 1-4 = -3. To find the geometric area, we must integrate the absolute value of the function, ∣x2−4∣|x^2 - 4|. Since x2−4≤0x^2 - 4 \le 0 on [0,2][0, 2], this is equivalent to integrating −(x2−4)-(x^2 - 4), which is 4−x24 - x^2. Therefore, the correct expression for the area is ∫02(4−x2)dx\int_0^2 (4-x^2) dx. Option A would give a negative result, which is the signed area, not the geometric area.

Question 8

Find ∫3x2+2x2dx\int \frac{3x^2+2}{x^2} dx.

  1. 3x+2x+C3x + \frac{2}{x} + C
  2. 3x−2x+C3x - \frac{2}{x} + C (correct answer)
  3. 3x−4x3+C3x - \frac{4}{x^3} + C
  4. ln⁡∣3x2+2∣−2ln⁡∣x∣+C\ln|3x^2+2| - 2\ln|x| + C
Explanation: Before integrating, simplify the fraction by splitting it into two terms: 3x2+2x2=3x2x2+2x2=3+2x−2\frac{3x^2+2}{x^2} = \frac{3x^2}{x^2} + \frac{2}{x^2} = 3 + 2x^{-2}. Now, integrate term by term: ∫(3+2x−2)dx=3x+2x−1−1+C=3x−2x−1+C=3x−2x+C\int (3 + 2x^{-2}) dx = 3x + \frac{2x^{-1}}{-1} + C = 3x - 2x^{-1} + C = 3x - \frac{2}{x} + C.

Question 9

A particle moves along a straight line with velocity v(t)=6t2−4t+1v(t) = 6t^2 - 4t + 1 m/s. Find the displacement of the particle between t=1t=1 and t=3t=3.

  1. 2 m
  2. 38 m (correct answer)
  3. 39 m
  4. 40 m
Explanation: Displacement is the definite integral of velocity over the time interval. We need to calculate ∫13(6t2−4t+1)dt\int_1^3 (6t^2 - 4t + 1) dt. First, find the antiderivative: ∫(6t2−4t+1)dt=6t33−4t22+t=2t3−2t2+t\int (6t^2 - 4t + 1) dt = \frac{6t^3}{3} - \frac{4t^2}{2} + t = 2t^3 - 2t^2 + t. Now evaluate this from t=1t=1 to t=3t=3: [2t3−2t2+t]13=(2(3)3−2(3)2+3)−(2(1)3−2(1)2+1)=(2(27)−2(9)+3)−(2−2+1)=(54−18+3)−(1)=39−1=38[2t^3 - 2t^2 + t]_1^3 = (2(3)^3 - 2(3)^2 + 3) - (2(1)^3 - 2(1)^2 + 1) = (2(27) - 2(9) + 3) - (2 - 2 + 1) = (54 - 18 + 3) - (1) = 39 - 1 = 38. The displacement is 38 m.

Question 10

Find ∫(2x−1x)2dx\int (2x - \frac{1}{x})^2 dx.

  1. 43x3+1x+C\frac{4}{3}x^3 + \frac{1}{x} + C
  2. 43x3−4x−1x+C\frac{4}{3}x^3 - 4x - \frac{1}{x} + C (correct answer)
  3. 43x3−1x+C\frac{4}{3}x^3 - \frac{1}{x} + C
  4. 13(2x−1x)3+C\frac{1}{3}(2x - \frac{1}{x})^3 + C
Explanation: First, expand the integrand: (2x−1x)2=(2x)2−2(2x)(1x)+(1x)2=4x2−4+1x2=4x2−4+x−2(2x - \frac{1}{x})^2 = (2x)^2 - 2(2x)(\frac{1}{x}) + (\frac{1}{x})^2 = 4x^2 - 4 + \frac{1}{x^2} = 4x^2 - 4 + x^{-2}. Now, integrate this expression term by term: ∫(4x2−4+x−2)dx=4x33−4x+x−1−1+C=43x3−4x−1x+C\int (4x^2 - 4 + x^{-2}) dx = \frac{4x^3}{3} - 4x + \frac{x^{-1}}{-1} + C = \frac{4}{3}x^3 - 4x - \frac{1}{x} + C.

Question 11

Given that ∫15f(x)dx=8\int_1^5 f(x) dx = 8 and ∫13f(x)dx=5\int_1^3 f(x) dx = 5, what is the value of ∫352f(x)dx\int_3^5 2f(x) dx?

  1. -2
  2. 3
  3. 6 (correct answer)
  4. 26
Explanation: Using the property of definite integrals, ∫acf(x)dx=∫abf(x)dx+∫bcf(x)dx\int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx. We have ∫15f(x)dx=∫13f(x)dx+∫35f(x)dx\int_1^5 f(x) dx = \int_1^3 f(x) dx + \int_3^5 f(x) dx. Substituting the given values: 8=5+∫35f(x)dx8 = 5 + \int_3^5 f(x) dx. This implies that ∫35f(x)dx=8−5=3\int_3^5 f(x) dx = 8 - 5 = 3. The question asks for ∫352f(x)dx\int_3^5 2f(x) dx, which is equal to 2∫35f(x)dx=2(3)=62\int_3^5 f(x) dx = 2(3) = 6.

Question 12

The gradient of a curve is given by f′(x)=ex/2−2xf'(x) = e^{x/2} - 2x. The curve passes through the point (0,5)(0, 5). Find the equation of the curve.

  1. y=ex/2−x2+4y = e^{x/2} - x^2 + 4
  2. y=2ex/2−x2+3y = 2e^{x/2} - x^2 + 3 (correct answer)
  3. y=12ex/2−x2+92y = \frac{1}{2}e^{x/2} - x^2 + \frac{9}{2}
  4. y=2ex/2−x2−3y = 2e^{x/2} - x^2 - 3
Explanation: To find the equation of the curve y=f(x)y = f(x), integrate the gradient function: f(x)=∫(ex/2−2x)dxf(x) = \int (e^{x/2} - 2x) dx. Using the reverse chain rule for ex/2e^{x/2} and the power rule for −2x-2x, we get f(x)=ex/21/2−2x22+C=2ex/2−x2+Cf(x) = \frac{e^{x/2}}{1/2} - \frac{2x^2}{2} + C = 2e^{x/2} - x^2 + C. To find CC, use the point (0,5)(0, 5): 5=2e0/2−02+C  ⟹  5=2e0+C  ⟹  5=2(1)+C  ⟹  C=35 = 2e^{0/2} - 0^2 + C \implies 5 = 2e^0 + C \implies 5 = 2(1) + C \implies C = 3. So the equation is y=2ex/2−x2+3y = 2e^{x/2} - x^2 + 3.

Question 13

Find the indefinite integral ∫e3−2xdx\int e^{3-2x} dx.

  1. −2e3−2x+C-2e^{3-2x} + C
  2. −12e3−2x+C-\frac{1}{2}e^{3-2x} + C (correct answer)
  3. 12e3−2x+C\frac{1}{2}e^{3-2x} + C
  4. e3−2x+Ce^{3-2x} + C
Explanation: This integral requires the reverse chain rule for a function of the form ∫f(ax+b)dx=1aF(ax+b)+C\int f(ax+b) dx = \frac{1}{a}F(ax+b) + C. Here, f(u)=euf(u)=e^u, and the inner function is 3−2x3-2x. The antiderivative of eue^u is eue^u. The derivative of the inner function is -2. Therefore, we must divide by -2. ∫e3−2xdx=1−2e3−2x+C=−12e3−2x+C\int e^{3-2x} dx = \frac{1}{-2}e^{3-2x} + C = -\frac{1}{2}e^{3-2x} + C.

Question 14

The area of the region enclosed by the curve y=3xy = \frac{3}{x}, the x-axis, and the lines x=1x=1 and x=e2x=e^2 is given by ∫1e23xdx\int_1^{e^2} \frac{3}{x} dx. Calculate this area.

  1. 2
  2. 3
  3. 6 (correct answer)
  4. 3e2−33e^2 - 3
Explanation: To find the area, we evaluate the definite integral. The antiderivative of 3x\frac{3}{x} is 3ln⁡∣x∣3\ln|x|. Using the Fundamental Theorem of Calculus: ∫1e23xdx=[3ln⁡∣x∣]1e2=3ln⁡(e2)−3ln⁡(1)\int_1^{e^2} \frac{3}{x} dx = [3\ln|x|]_1^{e^2} = 3\ln(e^2) - 3\ln(1). Using logarithm properties, ln⁡(e2)=2\ln(e^2) = 2 and ln⁡(1)=0\ln(1) = 0. So, the area is 3(2)−3(0)=63(2) - 3(0) = 6.

Question 15

If ∫f(x)dx=g(x)+C\int f(x) dx = g(x) + C, where CC is the constant of integration, which of the following statements must be true?

  1. f′(x)=g(x)f'(x) = g(x)
  2. g(x)=f(x)+Cg(x) = f(x) + C
  3. g′(x)=f(x)g'(x) = f(x) (correct answer)
  4. ∫g(x)dx=f(x)+C\int g(x) dx = f(x) + C
Explanation: The definition of an indefinite integral is that it is the antiderivative of the integrand. This means that if you differentiate the result of the integration, you get back the original function. Therefore, if ∫f(x)dx=g(x)+C\int f(x) dx = g(x) + C, then ddx(g(x)+C)=f(x)\frac{d}{dx}(g(x) + C) = f(x). Since the derivative of a constant CC is zero, this simplifies to g′(x)=f(x)g'(x) = f(x).

Question 16

Let f(x)=sin⁡(x3)f(x) = \sin(x^3). Which of the following definite integrals is equal to zero?

  1. ∫−π/2π/2f(x)dx\int_{-\pi/2}^{\pi/2} f(x) dx (correct answer)
  2. ∫0πf(x)dx\int_0^{\pi} f(x) dx
  3. ∫−ππ(f(x))2dx\int_{-\pi}^{\pi} (f(x))^2 dx
  4. ∫−π/2π/2(f(x)+1)dx\int_{-\pi/2}^{\pi/2} (f(x)+1) dx
Explanation: The function f(x)=sin⁡(x3)f(x) = \sin(x^3) is an odd function because f(−x)=sin⁡((−x)3)=sin⁡(−x3)=−sin⁡(x3)=−f(x)f(-x) = \sin((-x)^3) = \sin(-x^3) = -\sin(x^3) = -f(x). The definite integral of an odd function over a symmetric interval of the form [−a,a][-a, a] is always zero. Choice A, ∫−π/2π/2f(x)dx\int_{-\pi/2}^{\pi/2} f(x) dx, is an integral of an odd function over a symmetric interval, so its value is 0.

Question 17

Find ∫(x−2)2xdx\int \frac{(x-2)^2}{\sqrt{x}} dx.

  1. 25x5/2−83x3/2+8x+C\frac{2}{5}x^{5/2} - \frac{8}{3}x^{3/2} + 8\sqrt{x} + C (correct answer)
  2. 25x5/2+8x+C\frac{2}{5}x^{5/2} + 8\sqrt{x} + C
  3. 13(x−2)32x+C\frac{\frac{1}{3}(x-2)^3}{2\sqrt{x}} + C
  4. 25x5/2−2x3/2+4x+C\frac{2}{5}x^{5/2} - 2x^{3/2} + 4\sqrt{x} + C
Explanation: First, expand the numerator and rewrite the expression with fractional exponents: (x−2)2x=x2−4x+4x1/2\frac{(x-2)^2}{\sqrt{x}} = \frac{x^2 - 4x + 4}{x^{1/2}}. Then, divide each term in the numerator by x1/2x^{1/2}: x3/2−4x1/2+4x−1/2x^{3/2} - 4x^{1/2} + 4x^{-1/2}. Now, integrate term by term using the power rule ∫xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C: ∫(x3/2−4x1/2+4x−1/2)dx=x5/25/2−4x3/23/2+4x1/21/2+C\int (x^{3/2} - 4x^{1/2} + 4x^{-1/2}) dx = \frac{x^{5/2}}{5/2} - \frac{4x^{3/2}}{3/2} + \frac{4x^{1/2}}{1/2} + C. Simplifying gives: 25x5/2−83x3/2+8x1/2+C=25x5/2−83x3/2+8x+C\frac{2}{5}x^{5/2} - \frac{8}{3}x^{3/2} + 8x^{1/2} + C = \frac{2}{5}x^{5/2} - \frac{8}{3}x^{3/2} + 8\sqrt{x} + C.

Question 18

If ∫f(x)dx=g(x)+C\int f(x) dx = g(x) + C, where CC is the constant of integration, which of the following statements must be true?

  1. f′(x)=g(x)f'(x) = g(x)
  2. g(x)=f(x)+Cg(x) = f(x) + C
  3. g′(x)=f(x)g'(x) = f(x) (correct answer)
  4. ∫g(x)dx=f(x)+C\int g(x) dx = f(x) + C
Explanation: The definition of an indefinite integral is that it is the antiderivative of the integrand. This means that if you differentiate the result of the integration, you get back the original function. Therefore, if ∫f(x)dx=g(x)+C\int f(x) dx = g(x) + C, then ddx(g(x)+C)=f(x)\frac{d}{dx}(g(x) + C) = f(x). Since the derivative of a constant CC is zero, this simplifies to g′(x)=f(x)g'(x) = f(x).

Question 19

Find ∫cos⁡(4x−π2)dx\int \cos(4x - \frac{\pi}{2}) dx.

  1. −14sin⁡(4x−π2)+C-\frac{1}{4}\sin(4x - \frac{\pi}{2}) + C
  2. −4sin⁡(4x−π2)+C-4\sin(4x - \frac{\pi}{2}) + C
  3. 14sin⁡(4x−π2)+C\frac{1}{4}\sin(4x - \frac{\pi}{2}) + C (correct answer)
  4. sin⁡(4x−π2)+C\sin(4x - \frac{\pi}{2}) + C
Explanation: This requires the reverse chain rule. The integral of cos⁡(u)\cos(u) is sin⁡(u)\sin(u). The inner function is u=4x−π2u = 4x - \frac{\pi}{2}, and its derivative is u′=4u' = 4. We must divide the result by this derivative. Thus, ∫cos⁡(4x−π2)dx=14sin⁡(4x−π2)+C\int \cos(4x - \frac{\pi}{2}) dx = \frac{1}{4}\sin(4x - \frac{\pi}{2}) + C.

Question 20

Given that ∫15f(x)dx=8\int_1^5 f(x) dx = 8 and ∫13f(x)dx=5\int_1^3 f(x) dx = 5, what is the value of ∫352f(x)dx\int_3^5 2f(x) dx?

  1. -2
  2. 3
  3. 6 (correct answer)
  4. 26
Explanation: Using the property of definite integrals, ∫acf(x)dx=∫abf(x)dx+∫bcf(x)dx\int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx. We have ∫15f(x)dx=∫13f(x)dx+∫35f(x)dx\int_1^5 f(x) dx = \int_1^3 f(x) dx + \int_3^5 f(x) dx. Substituting the given values: 8=5+∫35f(x)dx8 = 5 + \int_3^5 f(x) dx. This implies that ∫35f(x)dx=8−5=3\int_3^5 f(x) dx = 8 - 5 = 3. The question asks for ∫352f(x)dx\int_3^5 2f(x) dx, which is equal to 2∫35f(x)dx=2(3)=62\int_3^5 f(x) dx = 2(3) = 6.