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

The integral from 0 to 1 of (e^x-2)^2 is 8+(e^2-1)/2-4e

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

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

8 + \frac{e ^{2} - 1}{2} - 4 e

Decimal Notation

0.32140 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{e} \frac{( u - 2 ) ^{2}}{u} d u

= \int_{1}^{e} \frac{( u - 2 ) ^{2}}{u} d u

Expand

\frac{( u - 2 ) ^{2}}{u} : u - 4 + \frac{4}{u}

= \int_{1}^{e} u - 4 + \frac{4}{u} d u

Apply the Sum Rule:

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

= \int_{1}^{e} u d u - \int_{1}^{e} 4 d u + \int_{1}^{e} \frac{4}{u} d u

\int_{1}^{e} u d u = \frac{e ^{2} - 1}{2}

\int_{1}^{e} 4 d u = 4 e - 4

\int_{1}^{e} \frac{4}{u} d u = 4

= \frac{e ^{2} - 1}{2} - (4 e - 4) + 4

Simplify

= 8 + \frac{e ^{2} - 1}{2} - 4 e

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What is the integral from 0 to 1 of (e^x-2)^2 ?
The integral from 0 to 1 of (e^x-2)^2 is 8+(e^2-1)/2-4e