What is the integral from 0 to 2 of x(e^{x^2}-e^4) ?

The integral from 0 to 2 of x(e^{x^2}-e^4) is (e^4-1)/2-e^4*2

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integral from 0 to 2 of x(e^{x^2}-e^4)

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

\frac{e ^{4} - 1}{2} - e^{4} \cdot 2

Decimal

- 82.39722 \dots

Solution steps

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

Expand

x (e^{x^{2}} - e^{4}) : e^{x^{2}} x - e^{4} x

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

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

\int_{0}^{2} e^{4} x d x = e^{4} \cdot 2

= \frac{e ^{4} - 1}{2} - e^{4} \cdot 2

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