What is the integral from 0 to 1 of (2e^x-5/(e^x))^2 ?

The integral from 0 to 1 of (2e^x-5/(e^x))^2 is 2e^2+(25(e^2-1))/(2e^2)-22

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integral from 0 to 1 of (2e^x-5/(e^x))^2

\int_{0}^{1} (2 e^{x} - \frac{5}{e ^{x}})^{2} d x

2 e^{2} + \frac{25 ( e ^{2} - 1 )}{2 e ^{2}} - 22

Decimal

3.58642 \dots

Solution steps

\int_{0}^{1} (2 e^{x} - \frac{5}{e ^{x}})^{2} d x

Simplify

(2 e^{x} - \frac{5}{e ^{x}})^{2} : 4 e^{2 x} - 20 + \frac{25}{e ^{2 x}}

= \int_{0}^{1} 4 e^{2 x} - 20 + \frac{25}{e ^{2 x}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} 4 e^{2 x} d x - \int_{0}^{1} 20 d x + \int_{0}^{1} \frac{25}{e ^{2 x}} d x

\int_{0}^{1} 4 e^{2 x} d x = 2 (e^{2} - 1)

\int_{0}^{1} 20 d x = 20

\int_{0}^{1} \frac{25}{e ^{2 x}} d x = \frac{25 ( e ^{2} - 1 )}{2 e ^{2}}

= 2 (e^{2} - 1) - 20 + \frac{25 ( e ^{2} - 1 )}{2 e ^{2}}

Simplify

2 (e^{2} - 1) - 20 + \frac{25 ( e ^{2} - 1 )}{2 e ^{2}} : 2 e^{2} + \frac{25 ( e ^{2} - 1 )}{2 e ^{2}} - 22

= 2 e^{2} + \frac{25 ( e ^{2} - 1 )}{2 e ^{2}} - 22

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What is the integral from 0 to 1 of (2e^x-5/(e^x))^2 ?
The integral from 0 to 1 of (2e^x-5/(e^x))^2 is 2e^2+(25(e^2-1))/(2e^2)-22