What is the integral from 0 to 2 of xe^{4x} ?

The integral from 0 to 2 of xe^{4x} is (7e^8+1)/(16)

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integral from 0 to 2 of xe^{4x}

\int_{0}^{2} x e^{4 x} d x

\frac{7 e ^{8} + 1}{16}

Decimal Notation

1304.23161 \dots

Solution steps

\int_{0}^{2} x e^{4 x} d x

Apply u-substitution

: \int_{0}^{8} \frac{e ^{u} u}{16} d u

= \int_{0}^{8} \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

= \frac{1}{16} \cdot \int_{0}^{8} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{8}

= \frac{1}{16} [e^{u} u - \int e^{u} d u]_{0}^{8}

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

= \frac{1}{16} [e^{u} u - e^{u}]_{0}^{8}

Compute the boundaries:

7 e^{8} + 1

= \frac{1}{16} (7 e^{8} + 1)

Simplify

= \frac{7 e ^{8} + 1}{16}

y^{^{'}} = y^{3}, y (0) = 1

\frac{d y}{d t} + \frac{1}{2} y = 17 sin (2 t)

\int cos (2 x) d x

\int (tan (x))^{4} d x

\frac{d}{d x} (\frac{x ^{2} - 5 x + 4}{∣ x - 1 ∣})

What is the integral from 0 to 2 of xe^{4x} ?
The integral from 0 to 2 of xe^{4x} is (7e^8+1)/(16)