Question

definite integral 1 and 0 5(e^x+1)^2 dx

Answer 1

To find the definite integral of the function 5(e^x+1)^2 from 0 to 1, we can use integration techniques.

First, we need to expand and simplify the expression (e^x+1)^2:

(e^x+1)^2 = (e^x+1)(e^x+1) = e^x * e^x + 2e^x + 1 = e^(2x) + 2e^x + 1

Now, we can integrate the simplified expression:

∫[0,1] 5(e^x+1)^2 dx = 5∫[0,1] (e^(2x) + 2e^x + 1) dx

Integrating term by term:

= 5(∫[0,1] e^(2x) dx + ∫[0,1] 2e^x dx + ∫[0,1] 1 dx)

Integrating each term separately:

∫ e^(2x) dx = (1/2)e^(2x) + C1

∫ 2e^x dx = 2e^x + C2

∫ 1 dx = x + C3

Where C1, C2, C3 are constants of integration.

Plugging in the limits:

= 5[((1/2)e^(2(1)) + C1) - ((1/2)e^(2(0)) + C1)] + 5[(2e^(1) + C2) - (2e^(0) + C2)] + 5[(1 + C3) - (0 + C3)]

Simplifying:

= 5[(1/2)e^2 + C1 - (1/2) - C1] + 5[2e + C2 - 2 - C2] + 5[1 + C3 - 0 - C3]

= 5[(1/2)e^2 - (1/2)] + 5[2e - 2] + 5[1]

= 5/2 * e^2 - 5/2 + 10e - 10 + 5

= 5/2 * e^2 + 10e - 5/2

Therefore, the definite integral of 5(e^x+1)^2 from 0 to 1 is 5/2 * e^2 + 10e - 5/2.