What is the integral from 0 to 1 of (7e^x-7xe^{x^2}) ?

The integral from 0 to 1 of (7e^x-7xe^{x^2}) is 7(e-1)-(7(e-1))/2

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integral from 0 to 1 of (7e^x-7xe^{x^2})

\int_{0}^{1} (7 e^{x} - 7 x e^{x^{2}}) d x

7 (e - 1) - \frac{7 ( e - 1 )}{2}

Decimal Notation

6.01398 \dots

Solution steps

\int_{0}^{1} 7 e^{x} - 7 x e^{x^{2}} 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} 7 e^{x} d x - \int_{0}^{1} 7 x e^{x^{2}} d x

\int_{0}^{1} 7 e^{x} d x = 7 (e - 1)

\int_{0}^{1} 7 x e^{x^{2}} d x = \frac{7 ( e - 1 )}{2}

= 7 (e - 1) - \frac{7 ( e - 1 )}{2}

f (x) = 3 x ln (19 x)

\int 3 x^{2} cos (2 x) d x

\frac{d y}{d x} - 2 x y = 4 x

\int 4 t^{3} - 8 t d t

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What is the integral from 0 to 1 of (7e^x-7xe^{x^2}) ?
The integral from 0 to 1 of (7e^x-7xe^{x^2}) is 7(e-1)-(7(e-1))/2