What is the integral from 1 to 5 of sqrt(2t-1) ?

The integral from 1 to 5 of sqrt(2t-1) is 26/3

Popular >

integral from 1 to 5 of sqrt(2t-1)

\int_{1}^{5} 2 t - 1 d t

\frac{26}{3}

Decimal Notation

8.66666 \dots

Solution steps

\int_{1}^{5} 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}^{4} (x (x - 1)) d x

\int_{0}^{4 π} ∣ sin (2 x) ∣ d x

\int_{0}^{\frac{π}{2}} x (- cos (x) + 1) d x

\int_{0}^{0.8} \frac{1}{( 1 - x ) ^{2}} d x

\int_{2}^{4} \frac{2 ( 1 + x )}{27} d x

What is the integral from 1 to 5 of sqrt(2t-1) ?
The integral from 1 to 5 of sqrt(2t-1) is 26/3