What is the integral from 0 to 1 of xe^{x^2}-e^x ?

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

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integral from 0 to 1 of xe^{x^2}-e^x

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

\frac{e - 1}{2} - e + 1

Decimal Notation

- 0.85914 \dots

Solution steps

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

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

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

= \frac{e - 1}{2} - (e - 1)

Simplify

= \frac{e - 1}{2} - e + 1

\int_{0}^{π} cos^{4} (x) d x

\frac{y ^{^{'}}}{x} = 2 x^{2} - y

f (x) = - \frac{4 x}{( x - 2 ) ^{3}}

g (t) = 5 - x

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