What is the integral from 2 to 6 of sqrt(1+1/4 x^1) ?

The integral from 2 to 6 of sqrt(1+1/4 x^1) is (2(5sqrt(10)-3sqrt(6)))/3

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integral from 2 to 6 of sqrt(1+1/4 x^1)

\int_{2}^{6} 1 + \frac{1}{4} x^{1} d x

\frac{2 ( 5 10 - 3 6 )}{3}

Decimal

5.64194 \dots

Solution steps

\int_{2}^{6} 1 + \frac{1}{4} x^{1} d x

Simplify

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

= \int_{2}^{6} \frac{1}{2} 4 + x d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{2}^{6} 4 + x d x

Apply u-substitution

: \int_{6}^{10} u d u

= \frac{1}{2} \cdot \int_{6}^{10} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{6}^{10}

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{6}^{10}

Compute the boundaries:

\frac{20 10}{3} - 46

= \frac{1}{2} (\frac{20 10}{3} - 46)

Simplify

= \frac{2 ( 5 10 - 3 6 )}{3}

f (x) = 3 x^{2} - 3 x + 4

\frac{\partial}{\partial x} (t^{4} + x^{3})

\int 4 cos (x) d x

\int \frac{1}{22 - x ^{2}} d x

n = 0 \sum \infty 6 (\frac{1}{7})^{n}

What is the integral from 2 to 6 of sqrt(1+1/4 x^1) ?
The integral from 2 to 6 of sqrt(1+1/4 x^1) is (2(5sqrt(10)-3sqrt(6)))/3