What is the integral from 1 to ln(2) of x*e^{4x} ?

The integral from 1 to ln(2) of x*e^{4x} is -(3e^4)/(16)+4ln(2)-1

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integral from 1 to ln(2) of x*e^{4x}

\int_{1}^{l n (2)} x \cdot e^{4 x} d x

- \frac{3 e ^{4}}{16} + 4 ln (2) - 1

Decimal Notation

- 8.46456 \dots

Solution steps

\int_{1}^{l n (2)} x e^{4 x} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{l n (2)}^{1} x e^{4 x} d x

Apply u-substitution

: \int_{4 l n (2)}^{4} \frac{e ^{u} u}{16} d u

= - \int_{4 l n (2)}^{4} \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_{4 l n (2)}^{4} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{4 l n (2)}^{4}

= - \frac{1}{16} [e^{u} u - \int e^{u} d u]_{4 l n (2)}^{4}

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

= - \frac{1}{16} [e^{u} u - e^{u}]_{4 l n (2)}^{4}

Compute the boundaries:

3 e^{4} - 64 ln (2) + 16

= - \frac{1}{16} (3 e^{4} - 64 ln (2) + 16)

Simplify

= - \frac{3 e ^{4}}{16} + 4 ln (2) - 1

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What is the integral from 1 to ln(2) of x*e^{4x} ?
The integral from 1 to ln(2) of x*e^{4x} is -(3e^4)/(16)+4ln(2)-1