What is the integral from 0 to 1 of ysqrt(3y+1) ?

The integral from 0 to 1 of ysqrt(3y+1) is 116/135

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integral from 0 to 1 of ysqrt(3y+1)

\int_{0}^{1} y 3 y + 1 d y

\frac{116}{135}

Decimal Notation

0.85925 \dots

Solution steps

\int_{0}^{1} y 3 y + 1 d y

Apply u-substitution

: \int_{1}^{4} \frac{u ( u - 1 )}{9} d u

= \int_{1}^{4} \frac{u ( u - 1 )}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{1}^{4} u (u - 1) d u

Expand

u (u - 1) : u^{\frac{3}{2}} - u

= \frac{1}{9} \cdot \int_{1}^{4} u^{\frac{3}{2}} - u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{9} (\int_{1}^{4} u^{\frac{3}{2}} d u - \int_{1}^{4} u d u)

\int_{1}^{4} u^{\frac{3}{2}} d u = \frac{62}{5}

\int_{1}^{4} u d u = \frac{14}{3}

= \frac{1}{9} (\frac{62}{5} - \frac{14}{3})

Simplify

\frac{1}{9} (\frac{62}{5} - \frac{14}{3}) : \frac{116}{135}

= \frac{116}{135}

\frac{d}{d x} ((5 - 3 x)^{2})

\int \frac{1}{( 1 + x ) ^{\frac{3}{4}}} d x

\int \frac{x ^{2} - x + 42}{x ^{3} + 7 x} d x

y = x^{\frac{1}{2}}

f (x) = 2 x + 5, at x = 2

What is the integral from 0 to 1 of ysqrt(3y+1) ?
The integral from 0 to 1 of ysqrt(3y+1) is 116/135