What is the integral from 0 to 4 of 3xe^{4x} ?

The integral from 0 to 4 of 3xe^{4x} is (3(15e^{16}+1))/(16)

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integral from 0 to 4 of 3xe^{4x}

\int_{0}^{4} 3 x e^{4 x} d x

\frac{3 ( 15 e ^{16} + 1 )}{16}

Decimal Notation

24992186.02642 \dots

Solution steps

\int_{0}^{4} 3 x e^{4 x} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{4} x e^{4 x} d x

Apply u-substitution

: \int_{0}^{16} \frac{e ^{u} u}{16} d u

= 3 \cdot \int_{0}^{16} \frac{e ^{u} u}{16} d u

Take the constant out:

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

= 3 \cdot \frac{1}{16} \cdot \int_{0}^{16} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{16}

= 3 \cdot \frac{1}{16} [e^{u} u - \int e^{u} d u]_{0}^{16}

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

= 3 \cdot \frac{1}{16} [e^{u} u - e^{u}]_{0}^{16}

Simplify

3 \cdot \frac{1}{16} [e^{u} u - e^{u}]_{0}^{16} : \frac{3}{16} [e^{u} u - e^{u}]_{0}^{16}

= \frac{3}{16} [e^{u} u - e^{u}]_{0}^{16}

Compute the boundaries:

15 e^{16} + 1

= \frac{3}{16} (15 e^{16} + 1)

Simplify

= \frac{3 ( 15 e ^{16} + 1 )}{16}

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What is the integral from 0 to 4 of 3xe^{4x} ?
The integral from 0 to 4 of 3xe^{4x} is (3(15e^{16}+1))/(16)