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

The integral from 0 to 1 of x^4e^{2x} is (8e^2-24)/(32)

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

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

\frac{8 e ^{2} - 24}{32}

Decimal Notation

1.09726 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{2} \frac{e ^{u} u ^{4}}{32} d u

= \int_{0}^{2} \frac{e ^{u} u ^{4}}{32} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{32} \cdot \int_{0}^{2} e^{u} u^{4} d u

Apply Integration By Parts

: [u^{4} e^{u} - \int 4 u^{3} e^{u} d u]_{0}^{2}

= \frac{1}{32} [u^{4} e^{u} - \int 4 u^{3} e^{u} d u]_{0}^{2}

\int 4 u^{3} e^{u} d u = 4 (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

= \frac{1}{32} [u^{4} e^{u} - 4 (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))]_{0}^{2}

Compute the boundaries:

8 e^{2} - 24

= \frac{1}{32} (8 e^{2} - 24)

Simplify

= \frac{8 e ^{2} - 24}{32}

\int \frac{1}{( x ^{2} + 6 ) ^{\frac{3}{2}}} d x

n = 0 \sum \infty 2 x^{n}

\frac{x}{x ^{2} + 36}

x \to 0 lim ((x)^{x})

y^{^{'}} = \frac{5 x}{y}

What is the integral from 0 to 1 of x^4e^{2x} ?
The integral from 0 to 1 of x^4e^{2x} is (8e^2-24)/(32)