What is the integral from 0 to 4 of sqrt(2x+1) ?

The integral from 0 to 4 of sqrt(2x+1) is 26/3

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integral from 0 to 4 of sqrt(2x+1)

\int_{0}^{4} 2 x + 1 d x

\frac{26}{3}

Decimal Notation

8.66666 \dots

Solution steps

\int_{0}^{4} 2 x + 1 d x

Apply u-substitution

: \int_{1}^{9} \frac{u}{2} d u

= \int_{1}^{9} \frac{u}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

Compute the boundaries:

\frac{52}{3}

= \frac{1}{2} \cdot \frac{52}{3}

Simplify

= \frac{26}{3}

x \to 3 + lim (\frac{6}{( x - 3 ) ^{5}})

y = \frac{4}{x ^{4}}

(x + 2)^{^{'}}

f (x) = (2 x^{3} + 3 x) (x - 3) (x + 5)

y^{^{''}} - 8 y^{^{'}} + 20 y = 100 x^{2} - 26 x e^{x}

What is the integral from 0 to 4 of sqrt(2x+1) ?
The integral from 0 to 4 of sqrt(2x+1) is 26/3