Question

Use the value of the first integral I to evaluate the two given integrals. IequalsIntegral from 0 to 1 (x cubed minus 5 x )dxequalsnegative nine fourths a. Integral from 0 to 1 (10 x minus 2 x cubed )dx b. Integral from 1 to 0 (5 x minus x cubed )dx

Answer 1

Answer:

a) (9/2)

b) (9/4)

Step-by-step explanation:

I = ∫¹₀ (x³ - 5x) dx = -(9/4)

a) ∫¹₀ (10x - 2x³) dx = -2 ∫¹₀ (x³ - 5x) dx

∫¹₀ (x³ - 5x) dx = -(9/4) from the given value for I

-2 ∫¹₀ (x³ - 5x) dx = -2 × (-9/4) = (9/2)

b) ∫¹₀ (5x - x³) dx = -1 ∫¹₀ (x³ - 5x) dx

∫¹₀ (x³ - 5x) dx = -(9/4) from the given value for I

-1 ∫¹₀ (x³ - 5x) dx = -1 × (-9/4) = (9/4)

Hope this Helps!!!

Answer 2

Answer:

a.

$\int\limits_{0}^{1} 10x - 2x^3 \,dx = -2(\int\limits_{0}^{1} x^3 -5x\,dx) = -2*(-9/4) = 9/2$

b.

$\int\limits_{0}^{1} 5x - x^3 \,dx = -1*(\int\limits_{0}^{1} x^3 -5x\,dx) = -1*(-9/4) = 9/4$

Step-by-step explanation:

According to the information given.

$\int\limits_{0}^{1} x^3 - 5x \,dx = -9/4$

Now.

a.

$\int\limits_{0}^{1} 10x - 2x^3 \,dx = -2(\int\limits_{0}^{1} x^3 -5x\,dx) = -2*(-9/4) = 9/2$

b.

$\int\limits_{0}^{1} 5x - x^3 \,dx = -1*(\int\limits_{0}^{1} x^3 -5x\,dx) = -1*(-9/4) = 9/4$