What is the integral from 0 to 1 of 8xe^{9x} ?

The integral from 0 to 1 of 8xe^{9x} is (8(8e^9+1))/(81)

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integral from 0 to 1 of 8xe^{9x}

\int_{0}^{1} 8 x e^{9 x} d x

\frac{8 ( 8 e ^{9} + 1 )}{81}

Decimal

6402.53544 \dots

Solution steps

\int_{0}^{1} 8 x e^{9 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int_{0}^{1} x e^{9 x} d x

Apply u-substitution

= 8 \cdot \int_{0}^{9} \frac{e ^{u} u}{81} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \frac{1}{81} \cdot \int_{0}^{9} e^{u} u d u

Apply Integration By Parts

= 8 \cdot \frac{1}{81} [e^{u} u - \int e^{u} d u]_{0}^{9}

\int e^{u} d u = e^{u}

= 8 \cdot \frac{1}{81} [e^{u} u - e^{u}]_{0}^{9}

Simplify

8 \cdot \frac{1}{81} [e^{u} u - e^{u}]_{0}^{9} : \frac{8}{81} [e^{u} u - e^{u}]_{0}^{9}

= \frac{8}{81} [e^{u} u - e^{u}]_{0}^{9}

Compute the boundaries:

8 e^{9} + 1

= \frac{8}{81} (8 e^{9} + 1)

Simplify

= \frac{8 ( 8 e ^{9} + 1 )}{81}

What is the integral from 0 to 1 of 8xe^{9x} ?
The integral from 0 to 1 of 8xe^{9x} is (8(8e^9+1))/(81)