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

The integral from 0 to 3 of xe^{4x} is (11e^{12}+1)/(16)

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

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

\frac{11 e ^{12} + 1}{16}

Decimal Notation

111893.98160 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{12} \frac{e ^{u} u}{16} d u

= \int_{0}^{12} \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}^{12} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{12}

= \frac{1}{16} [e^{u} u - \int e^{u} d u]_{0}^{12}

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

= \frac{1}{16} [e^{u} u - e^{u}]_{0}^{12}

Compute the boundaries:

11 e^{12} + 1

= \frac{1}{16} (11 e^{12} + 1)

Simplify

= \frac{11 e ^{12} + 1}{16}

3 sin (t)

x \to 0 lim (\frac{3 x}{x})

f (x) = \frac{2}{x}, (- 6, - \frac{1}{3})

\frac{d y}{d x} = 4 y^{3}

(2 x y + y) d x + (x^{2} - x) d y = 0

What is the integral from 0 to 3 of xe^{4x} ?
The integral from 0 to 3 of xe^{4x} is (11e^{12}+1)/(16)