What is the integral from 0 to 1 of x^4e^{3x} ?

The integral from 0 to 1 of x^4e^{3x} is (11e^3-8)/(81)

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integral from 0 to 1 of x^4e^{3x}

\int_{0}^{1} x^{4} e^{3 x} d x

\frac{11 e ^{3} - 8}{81}

Decimal Notation

2.62890 \dots

Solution steps

\int_{0}^{1} x^{4} e^{3 x} d x

Apply Integration By Parts

: [\frac{1}{3} e^{3 x} x^{4} - \int \frac{4}{3} e^{3 x} x^{3} d x]_{0}^{1}

= [\frac{1}{3} e^{3 x} x^{4} - \int \frac{4}{3} e^{3 x} x^{3} d x]_{0}^{1}

\int \frac{4}{3} e^{3 x} x^{3} d x = \frac{4}{81} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2)

= [\frac{1}{3} e^{3 x} x^{4} - \frac{4}{81} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2)]_{0}^{1}

Compute the boundaries:

\frac{11 e ^{3}}{81} - \frac{8}{81}

= \frac{11 e ^{3}}{81} - \frac{8}{81}

Simplify

= \frac{11 e ^{3} - 8}{81}

y = \frac{e ^{x}}{x}, (1, e)

\int ((sin (x))^{4} (cos (x))^{2}) d x

\frac{2}{x}

\int e^{x} sin (e^{x}) cos (e^{x}) d x

x - cos (5 y) = 0

What is the integral from 0 to 1 of x^4e^{3x} ?
The integral from 0 to 1 of x^4e^{3x} is (11e^3-8)/(81)