What is the integral from 0 to 2 of 10x^3sqrt(x^2+4) ?

The integral from 0 to 2 of 10x^3sqrt(x^2+4) is (2(64sqrt(2)+64))/3

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integral from 0 to 2 of 10x^3sqrt(x^2+4)

\int_{0}^{2} 10 x^{3} x^{2} + 4 d x

\frac{2 ( 64 2 + 64 )}{3}

Decimal Notation

103.00644 \dots

Solution steps

\int_{0}^{2} 10 x^{3} x^{2} + 4 d x

Take the constant out:

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

= 10 \cdot \int_{0}^{2} x^{3} x^{2} + 4 d x

Apply u-substitution

: \int_{2}^{22} (u^{2} - 4) u^{2} d u

= 10 \cdot \int_{2}^{22} (u^{2} - 4) u^{2} d u

Expand

(u^{2} - 4) u^{2} : u^{4} - 4 u^{2}

= 10 \cdot \int_{2}^{22} u^{4} - 4 u^{2} d u

Apply the Sum Rule:

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

= 10 (\int_{2}^{22} u^{4} d u - \int_{2}^{22} 4 u^{2} d u)

\int_{2}^{22} u^{4} d u = \frac{128 2 - 32}{5}

\int_{2}^{22} 4 u^{2} d u = \frac{4 ( 16 2 - 8 )}{3}

= 10 (\frac{128 2 - 32}{5} - \frac{4 ( 16 2 - 8 )}{3})

Simplify

10 (\frac{128 2 - 32}{5} - \frac{4 ( 16 2 - 8 )}{3}) : \frac{2 ( 64 2 + 64 )}{3}

= \frac{2 ( 64 2 + 64 )}{3}

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 3 y = 2 x - 17

\frac{\partial}{\partial y} (x tan (y))

y = - 2 x^{2} - x + 6

\int_{0}^{a} x sin^{2} (\frac{π}{a} x) d x

\frac{\partial}{\partial x} (- 2 x y^{2})

What is the integral from 0 to 2 of 10x^3sqrt(x^2+4) ?
The integral from 0 to 2 of 10x^3sqrt(x^2+4) is (2(64sqrt(2)+64))/3