What is the integral from 0 to 1 of x^4e^{6x} ?

The integral from 0 to 1 of x^4e^{6x} is (31e^6-1)/(324)

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integral from 0 to 1 of x^4e^{6x}

\int_{0}^{1} x^{4} e^{6 x} d x

\frac{31 e ^{6} - 1}{324}

Decimal Notation

38.59658 \dots

Solution steps

\int_{0}^{1} x^{4} e^{6 x} d x

Apply Integration By Parts

: [\frac{1}{6} e^{6 x} x^{4} - \int \frac{2}{3} e^{6 x} x^{3} d x]_{0}^{1}

= [\frac{1}{6} e^{6 x} x^{4} - \int \frac{2}{3} e^{6 x} x^{3} d x]_{0}^{1}

\int \frac{2}{3} e^{6 x} x^{3} d x = \frac{1}{324} e^{6 x} (36 x^{3} - 18 x^{2} + 6 x - 1)

= [\frac{1}{6} e^{6 x} x^{4} - \frac{1}{324} e^{6 x} (36 x^{3} - 18 x^{2} + 6 x - 1)]_{0}^{1}

Compute the boundaries:

\frac{31 e ^{6}}{324} - \frac{1}{324}

= \frac{31 e ^{6}}{324} - \frac{1}{324}

Simplify

= \frac{31 e ^{6} - 1}{324}

\int - t z d t

e^{2 x} y^{^{'}} - 2 e^{2 x} y = 3 e^{4 x}

x \to 5 - lim (\frac{12}{5 - x})

\int - 4 sin (3 x) + 6 sin (8 x) d x

\frac{d}{d x} (e^{x} (8 x^{2} - 16 x + 16))

What is the integral from 0 to 1 of x^4e^{6x} ?
The integral from 0 to 1 of x^4e^{6x} is (31e^6-1)/(324)