Assume that we have functions f(x) and g(x) so that ∫−13f(x)dx=−1,∫35f(x)dx=4,∫−11g(x)dx=3,∫15g(x)dx=2 Determine ∫−15(2f(x)−5g(x))dx.

Step 1 Given: ∫−13f(x)dx=−1 ∫35f(x)dx=4 ∫−11g(x)dx=3 ∫15g(x)dx=2 To find: ∫−15(2f(x)−5g(x))dx Step 2 Using difference rule, we get: ⇒∫−15(2f(x)−5g(x))dx=∫−15(2f(x))dx−∫−15(5g(x))dx the constant can be taken out of the integral, hence we get: ⇒∫−15(2f(x)−5g(x))dx=2∫−15f(x)dx−5∫−15g(x)dx Step 3 splitting the above integrals, we get: ⇒∫−15(2f(x)−5g(x))dx=2(∫−13f(x)dx+∫35f(x)dx) −5(∫−11g(x)dx+∫15g(x)dx) plugging in the given values of these integrals, we get: ⇒∫−15(2f(x)−5g(x))dx=2(−1+4)−5(3+2)

Expert Answer to "Assume that we have functions f(x) and g(x)..."

shadsiei

Answered question

2021-02-14

Assume that we have functions f(x) and g(x) so that

\int_{- 1}^{3} f (x) d x = - 1, \int_{3}^{5} f (x) d x = 4, \int_{- 1}^{1} g (x) d x = 3, \int_{1}^{5} g (x) d x = 2

Determine

\int_{- 1}^{5} (2 f (x) - 5 g (x)) d x

Answer & Explanation

Demi-Leigh Barrera

Skilled2021-02-15Added 97 answers

Step 1
Given:

\int_{- 1}^{3} f (x) d x = - 1

\int_{3}^{5} f (x) d x = 4

\int_{- 1}^{1} g (x) d x = 3

\int_{1}^{5} g (x) d x = 2

To find:

\int_{- 1}^{5} (2 f (x) - 5 g (x)) d x

Step 2
Using difference rule, we get:

\Rightarrow \int_{- 1}^{5} (2 f (x) - 5 g (x)) d x = \int_{- 1}^{5} (2 f (x)) d x - \int_{- 1}^{5} (5 g (x)) d x

the constant can be taken out of the integral, hence we get:

\Rightarrow \int_{- 1}^{5} (2 f (x) - 5 g (x)) d x = 2 \int_{- 1}^{5} f (x) d x - 5 \int_{- 1}^{5} g (x) d x

Step 3
splitting the above integrals, we get:

\Rightarrow \int_{- 1}^{5} (2 f (x) - 5 g (x)) d x = 2 (\int_{- 1}^{3} f (x) d x + \int_{3}^{5} f (x) d x)

- 5 (\int_{- 1}^{1} g (x) d x + \int_{1}^{5} g (x) d x)

plugging in the given values of these integrals, we get:

\Rightarrow \int_{- 1}^{5} (2 f (x) - 5 g (x)) d x = 2 (- 1 + 4) - 5 (3 + 2)

This is helpful 34 Flag

Jeffrey Jordon

Expert2021-11-05Added 2605 answers

Answer is given below (on video)

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Integral Calculus

Didn't find what you were looking for?