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

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

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

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

\frac{26}{3}

Decimal Notation

8.66666 \dots

Solution steps

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

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}

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

\int_{1}^{3} \frac{1}{x ^{2} + 2 x} d x

\int_{1}^{2} (x) d x

\int_{0}^{π} e^{x} d x

\int_{x}^{1} sin (y^{2}) d y

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