What is the integral from 2 to 4 of (x^2+2)^3 ?

The integral from 2 to 4 of (x^2+2)^3 is 131344/35

integral from 2 to 4 of (x^2+2)^3

\int_{2}^{4} (x^{2} + 2)^{3} d x

\frac{131344}{35}

Decimal

3752.68571 \dots

Solution steps

\int_{2}^{4} (x^{2} + 2)^{3} d x

Expand

(x^{2} + 2)^{3} : x^{6} + 6 x^{4} + 12 x^{2} + 8

= \int_{2}^{4} x^{6} + 6 x^{4} + 12 x^{2} + 8 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_{2}^{4} x^{6} d x + \int_{2}^{4} 6 x^{4} d x + \int_{2}^{4} 12 x^{2} d x + \int_{2}^{4} 8 d x

\int_{2}^{4} x^{6} d x = \frac{16256}{7}

\int_{2}^{4} 6 x^{4} d x = \frac{5952}{5}

\int_{2}^{4} 12 x^{2} d x = 224

\int_{2}^{4} 8 d x = 16

= \frac{16256}{7} + \frac{5952}{5} + 224 + 16

Simplify

\frac{16256}{7} + \frac{5952}{5} + 224 + 16 : \frac{131344}{35}

= \frac{131344}{35}

y^{^{''}} + y = t, y (0) = 1, y (1) = - 2

\frac{d}{d x} (\frac{1}{2 x} + \frac{1}{8 x ^{\frac{7}{8}}})

l a pl a ce t r an s f or m e^{- 2 t} (2 cos (3 t) - sin (3 t))

x \to \infty lim (\frac{5}{x})

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

What is the integral from 2 to 4 of (x^2+2)^3 ?
The integral from 2 to 4 of (x^2+2)^3 is 131344/35