What is the integral from 0 to 1 of 4xe^{7x} ?

The integral from 0 to 1 of 4xe^{7x} is (4(6e^7+1))/(49)

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integral from 0 to 1 of 4xe^{7x}

\int_{0}^{1} 4 x e^{7 x} d x

\frac{4 ( 6 e ^{7} + 1 )}{49}

Decimal

537.20807 \dots

Solution steps

\int_{0}^{1} 4 x e^{7 x} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{1} x e^{7 x} d x

Apply u-substitution

= 4 \cdot \int_{0}^{7} \frac{e ^{u} u}{49} d u

Take the constant out:

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

= 4 \cdot \frac{1}{49} \cdot \int_{0}^{7} e^{u} u d u

Apply Integration By Parts

= 4 \cdot \frac{1}{49} [e^{u} u - \int e^{u} d u]_{0}^{7}

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

= 4 \cdot \frac{1}{49} [e^{u} u - e^{u}]_{0}^{7}

Simplify

4 \cdot \frac{1}{49} [e^{u} u - e^{u}]_{0}^{7} : \frac{4}{49} [e^{u} u - e^{u}]_{0}^{7}

= \frac{4}{49} [e^{u} u - e^{u}]_{0}^{7}

Compute the boundaries:

6 e^{7} + 1

= \frac{4}{49} (6 e^{7} + 1)

Simplify

= \frac{4 ( 6 e ^{7} + 1 )}{49}

What is the integral from 0 to 1 of 4xe^{7x} ?
The integral from 0 to 1 of 4xe^{7x} is (4(6e^7+1))/(49)