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

The integral from 1 to 2 of 4xsqrt(4x^2+4) is (8(5sqrt(5)-2sqrt(2)))/3

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

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

\frac{8 ( 5 5 - 2 2 )}{3}

Decimal Notation

22.27176 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{8}^{20} \frac{u}{8} d u

= 4 \cdot \int_{8}^{20} \frac{u}{8} d u

Take the constant out:

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

= 4 \cdot \frac{1}{8} \cdot \int_{8}^{20} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{20}

= 4 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{20}

Simplify

4 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{20} : \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{20}

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{20}

Compute the boundaries:

\frac{80 5 - 32 2}{3}

= \frac{1}{2} \cdot \frac{80 5 - 32 2}{3}

Simplify

= \frac{8 ( 5 5 - 2 2 )}{3}

\int (3 x^{4} - 5 x^{2} + 6 x) d x

x \to 0 lim (\frac{∣ x + 1 ∣ - ∣ x - 1 ∣}{x})

n = 1 \sum \infty 8

\frac{\partial}{\partial v} (u^{2} + v^{2})

1 - e^{- 0.4 t}

What is the integral from 1 to 2 of 4xsqrt(4x^2+4) ?
The integral from 1 to 2 of 4xsqrt(4x^2+4) is (8(5sqrt(5)-2sqrt(2)))/3