What is the integral from 0 to 1 of 80x(2x^2+1)^4 ?

The integral from 0 to 1 of 80x(2x^2+1)^4 is 968

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 80x(2x^2+1)^4

\int_{0}^{1} 80 x (2 x^{2} + 1)^{4} d x

968

Solution steps

\int_{0}^{1} 80 x (2 x^{2} + 1)^{4} d x

Take the constant out:

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

= 80 \cdot \int_{0}^{1} x (2 x^{2} + 1)^{4} d x

Multiply by the conjugate of

2 x^{2} + 1 : \frac{4 x ^{4} - 1}{2 x ^{2} - 1}

= 80 \cdot \int_{0}^{1} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x

Find undefined (singularity) points:

x = \frac{1}{2}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= 80 (\int_{0}^{\frac{1}{2}} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x + \int_{\frac{1}{2}}^{1} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x)

\int_{0}^{\frac{1}{2}} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x = \frac{31}{20}

\int_{\frac{1}{2}}^{1} x (\frac{4 x ^{4} - 1}{2 x ^{2} - 1})^{4} d x = \frac{211}{20}

= 80 (\frac{31}{20} + \frac{211}{20})

Simplify

80 (\frac{31}{20} + \frac{211}{20}) : 968

= 968

\int_{- π}^{0} - cos (n x) d x

\int_{0}^{1} x e^{- s x} d x

\int_{0}^{1} ln (y) d y

\int_{0}^{- π} cos (x) d x

\int_{0}^{y} \frac{6 ( y - x )}{125} d x

What is the integral from 0 to 1 of 80x(2x^2+1)^4 ?
The integral from 0 to 1 of 80x(2x^2+1)^4 is 968