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

The integral from-1/2 to 0 of x^4*e^{2x+1} is (24e-65)/(32)

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integral from-1/2 to 0 of x^4*e^{2x+1}

\int_{- \frac{1}{2}}^{0} x^{4} \cdot e^{2 x + 1} d x

\frac{24 e - 65}{32}

Decimal

0.00746 \dots

Solution steps

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

x^{4} e^{2 x + 1} = e^{1} x^{4} e^{2 x}

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

Take the constant out:

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

= e^{1} \cdot \int_{- \frac{1}{2}}^{0} x^{4} e^{2 x} d x

Simplify

= e \cdot \int_{- \frac{1}{2}}^{0} x^{4} e^{2 x} d x

Apply u-substitution

= e \cdot \int_{- 1}^{0} \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

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

Apply Integration By Parts

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

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

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

Simplify

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

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

Compute the boundaries:

24 - \frac{65}{e}

= \frac{e}{32} (24 - \frac{65}{e})

Simplify

= \frac{24 e - 65}{32}

What is the integral from-1/2 to 0 of x^4*e^{2x+1} ?
The integral from-1/2 to 0 of x^4*e^{2x+1} is (24e-65)/(32)