What is the integral from 0 to 1 of sqrt(1+9/4 y) ?

The integral from 0 to 1 of sqrt(1+9/4 y) is (13sqrt(13)-8)/(27)

Popular >

integral from 0 to 1 of sqrt(1+9/4 y)

\int_{0}^{1} 1 + \frac{9}{4} y d y

\frac{13 13 - 8}{27}

Decimal Notation

1.43970 \dots

Solution steps

\int_{0}^{1} 1 + \frac{9}{4} y d y

Apply u-substitution

: \int_{1}^{\frac{13}{4}} \frac{4 u}{9} d u

= \int_{1}^{\frac{13}{4}} \frac{4 u}{9} d u

Take the constant out:

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

= \frac{4}{9} \cdot \int_{1}^{\frac{13}{4}} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{\frac{13}{4}}

= \frac{4}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{\frac{13}{4}}

Compute the boundaries:

\frac{13 13}{12} - \frac{2}{3}

= \frac{4}{9} (\frac{13 13}{12} - \frac{2}{3})

Simplify

= \frac{13 13 - 8}{27}

\int_{1}^{3} x^{5} x^{6} - 1 d x

\int_{0}^{\frac{π}{3}} 6 x sin (x) d x

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

\int_{0}^{t} (t - x)^{3} e^{t} d x

\int_{0}^{1} \frac{x}{x - 1} d x

What is the integral from 0 to 1 of sqrt(1+9/4 y) ?
The integral from 0 to 1 of sqrt(1+9/4 y) is (13sqrt(13)-8)/(27)