What is the integral from-1 to 1 of |e^{2x}-10| ?

The integral from-1 to 1 of |e^{2x}-10| is -(e^4-1)/(2e^2)+20

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integral from-1 to 1 of |e^{2x}-10|

\int_{- 1}^{1} e^{2 x} - 10 d x

- \frac{e ^{4} - 1}{2 e ^{2}} + 20

Decimal

16.37313 \dots

Solution steps

\int_{- 1}^{1} e^{2 x} - 10 d x

Eliminate Absolutes

= \int_{- 1}^{1} - e^{2 x} + 10 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_{- 1}^{1} e^{2 x} d x + \int_{- 1}^{1} 10 d x

\int_{- 1}^{1} e^{2 x} d x = \frac{e ^{4} - 1}{2 e ^{2}}

\int_{- 1}^{1} 10 d x = 20

= - \frac{e ^{4} - 1}{2 e ^{2}} + 20

What is the integral from-1 to 1 of |e^{2x}-10| ?
The integral from-1 to 1 of |e^{2x}-10| is -(e^4-1)/(2e^2)+20