What is the integral from 1 to 2 of (x^4)/8+1/(4x^2) ?

The integral from 1 to 2 of (x^4)/8+1/(4x^2) is 9/10

Popular >

integral from 1 to 2 of (x^4)/8+1/(4x^2)

\int_{1}^{2} \frac{x ^{4}}{8} + \frac{1}{4 x ^{2}} d x

\frac{9}{10}

Decimal Notation

0.9

Solution steps

\int_{1}^{2} \frac{x ^{4}}{8} + \frac{1}{4 x ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} \frac{x ^{4}}{8} d x + \int_{1}^{2} \frac{1}{4 x ^{2}} d x

\int_{1}^{2} \frac{x ^{4}}{8} d x = \frac{31}{40}

\int_{1}^{2} \frac{1}{4 x ^{2}} d x = \frac{1}{8}

= \frac{31}{40} + \frac{1}{8}

Simplify

\frac{31}{40} + \frac{1}{8} : \frac{9}{10}

= \frac{9}{10}

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

(\frac{( x + 1 )}{( x + 1 ) ^{2} + 4})^{^{'}}

\frac{d}{d x} (x^{5} sin (3 x))

\frac{d}{d x} ((x^{5} - \frac{6}{x})^{- 2})

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

What is the integral from 1 to 2 of (x^4)/8+1/(4x^2) ?
The integral from 1 to 2 of (x^4)/8+1/(4x^2) is 9/10