Math Problem Statement

Solve

Suppose that f and g are integrable functions and that ∫[1,2] f(x) dx = -4, ∫[1,5] f(x) dx = 6, and ∫[1,5] g(x) dx = 8. Find the following integrals: (a) ∫[3,5] 2g(x) dx (b) ∫[5,3] g(x) dx (c) ∫[1,2] 3f(x) dx (d) ∫[2,5] f(x) dx (e) ∫[1,5] [g(x) - f(x)] dx (f) ∫[1,5] [4f(x) - g(x)] dx

Solution

Let's solve each part step-by-step using the given integrals:

Given:

  1. 12f(x)dx=4\int_{1}^{2} f(x) \, dx = -4
  2. 15f(x)dx=6\int_{1}^{5} f(x) \, dx = 6
  3. 15g(x)dx=8\int_{1}^{5} g(x) \, dx = 8

We will use these values to find each integral in the question.

Solution for each part:

Part (a)

352g(x)dx\int_{3}^{5} 2g(x) \, dx Using the property of integrals, we can factor out constants: 352g(x)dx=235g(x)dx\int_{3}^{5} 2g(x) \, dx = 2 \int_{3}^{5} g(x) \, dx To evaluate 35g(x)dx\int_{3}^{5} g(x) \, dx, we use: 15g(x)dx=13g(x)dx+35g(x)dx\int_{1}^{5} g(x) \, dx = \int_{1}^{3} g(x) \, dx + \int_{3}^{5} g(x) \, dx So, 8=13g(x)dx+35g(x)dx8 = \int_{1}^{3} g(x) \, dx + \int_{3}^{5} g(x) \, dx Let 35g(x)dx=I\int_{3}^{5} g(x) \, dx = I. Then, 8=13g(x)dx+I8 = \int_{1}^{3} g(x) \, dx + I Since we do not have 13g(x)dx\int_{1}^{3} g(x) \, dx, we cannot solve this part directly without further information.

Part (b)

53g(x)dx\int_{5}^{3} g(x) \, dx Using the property of reversing limits in integrals: 53g(x)dx=35g(x)dx\int_{5}^{3} g(x) \, dx = -\int_{3}^{5} g(x) \, dx Without the value of 35g(x)dx\int_{3}^{5} g(x) \, dx, we cannot solve this directly either.

Part (c)

123f(x)dx\int_{1}^{2} 3f(x) \, dx Using the constant multiple rule: 123f(x)dx=312f(x)dx=3(4)=12\int_{1}^{2} 3f(x) \, dx = 3 \int_{1}^{2} f(x) \, dx = 3 \cdot (-4) = -12

Part (d)

25f(x)dx\int_{2}^{5} f(x) \, dx We know: 15f(x)dx=12f(x)dx+25f(x)dx\int_{1}^{5} f(x) \, dx = \int_{1}^{2} f(x) \, dx + \int_{2}^{5} f(x) \, dx So, 6=4+25f(x)dx6 = -4 + \int_{2}^{5} f(x) \, dx 25f(x)dx=6+4=10\int_{2}^{5} f(x) \, dx = 6 + 4 = 10

Part (e)

15[g(x)f(x)]dx\int_{1}^{5} \left[g(x) - f(x)\right] \, dx Using the linearity of integrals: 15[g(x)f(x)]dx=15g(x)dx15f(x)dx\int_{1}^{5} \left[g(x) - f(x)\right] \, dx = \int_{1}^{5} g(x) \, dx - \int_{1}^{5} f(x) \, dx =86=2= 8 - 6 = 2

Part (f)

15[4f(x)g(x)]dx\int_{1}^{5} \left[4f(x) - g(x)\right] \, dx Again, using linearity: 15[4f(x)g(x)]dx=415f(x)dx15g(x)dx\int_{1}^{5} \left[4f(x) - g(x)\right] \, dx = 4 \int_{1}^{5} f(x) \, dx - \int_{1}^{5} g(x) \, dx =468=248=16= 4 \cdot 6 - 8 = 24 - 8 = 16

Summary of Answers

  • (a) Requires additional information for 35g(x)dx\int_{3}^{5} g(x) \, dx.
  • (b) Depends on the result of (a).
  • (c) 12-12
  • (d) 1010
  • (e) 22
  • (f) 1616

Would you like details on any specific part, or do you have any questions?


Here are five related questions that expand on this information:

  1. How would the answers change if 13g(x)dx\int_{1}^{3} g(x) \, dx were given?
  2. Can we derive 13g(x)dx\int_{1}^{3} g(x) \, dx using any additional assumptions?
  3. How do properties of linearity help simplify complex integral expressions?
  4. What methods could be used if we need numerical values for unknown integrals?
  5. How can we verify these results if given functions f(x)f(x) and g(x)g(x) explicitly?

Tip: When given partial information on integrals over different intervals, always check if adding or subtracting known values can help fill in the gaps.

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Math Problem Analysis

Mathematical Concepts

Definite Integrals
Properties of Integrals
Linearity of Integrals

Formulas

∫[a,b] k * f(x) dx = k * ∫[a,b] f(x) dx
∫[a,b] [f(x) ± g(x)] dx = ∫[a,b] f(x) dx ± ∫[a,b] g(x) dx
∫[a,b] f(x) dx = -∫[b,a] f(x) dx

Theorems

Properties of Definite Integrals

Suitable Grade Level

Grades 11-12