What is the integral from 0 to 1 of x^2e^{4x} ?

The integral from 0 to 1 of x^2e^{4x} is (5e^4-1)/(32)

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integral from 0 to 1 of x^2e^{4x}

\int_{0}^{1} x^{2} e^{4 x} d x

\frac{5 e ^{4} - 1}{32}

Decimal Notation

8.49971 \dots

Solution steps

\int_{0}^{1} x^{2} e^{4 x} d x

Apply Integration By Parts

: [\frac{1}{4} e^{4 x} x^{2} - \int \frac{1}{2} e^{4 x} x d x]_{0}^{1}

= [\frac{1}{4} e^{4 x} x^{2} - \int \frac{1}{2} e^{4 x} x d x]_{0}^{1}

\int \frac{1}{2} e^{4 x} x d x = \frac{1}{32} (4 e^{4 x} x - e^{4 x})

= [\frac{1}{4} e^{4 x} x^{2} - \frac{1}{32} (4 e^{4 x} x - e^{4 x})]_{0}^{1}

Compute the boundaries:

\frac{5 e ^{4}}{32} - \frac{1}{32}

= \frac{5 e ^{4}}{32} - \frac{1}{32}

Simplify

= \frac{5 e ^{4} - 1}{32}

n = 1 \sum \infty \frac{1}{4 ^{n} - 3 ^{n}}

2 (2 x y + 4 y - 3) d x + (x + 2)^{2} d y = 0

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n = 1 \sum \infty \frac{3 n}{n ^{2} + 2}

\int_{0}^{2 π} 1

What is the integral from 0 to 1 of x^2e^{4x} ?
The integral from 0 to 1 of x^2e^{4x} is (5e^4-1)/(32)