MHB How Do You Evaluate the Integral of y² / √(4-3y) from 0 to 1?

[karush]

$\int_{0}^{1} \frac{y^2}{\sqrt{4-3y}}dy$

I know there is several ways to substitute this but I chose

$u=4-3y, du=3 dy, y^2=u-4$

The answer was $\frac{106}{405}$

But I couldn't get it.

 

[MarkFL]

Using your substitution of:

$$u=4-3y$$

we would have:

$$du=-3\,dy$$

$$y^2=\frac{(4-u)^2}{9}$$

And so the definite integral becomes:

$$I=\frac{1}{27}\int_1^4\frac{(4-u)^2}{\sqrt{u}}\,du$$

Can you proceed?

 

[karush]

$\frac{1}{27}
\int_{0}^{4} \left(\frac{16}{\sqrt{u}}
-\frac{8u}{\sqrt{u}}
+\frac{{u}^{2}}{\sqrt{u}}\right)\,du$

I just expanded the denominator and parted out the fraction hope this is correct. Why is there a 4 in the interval?

 

[MarkFL]

When you make the substitution, everything in terms of $y$, the integrand, the differential and the limits of integration, need to be changed in accordance with the substitution. So that's why the limits changed. I would then simplify the integral to:

$$I=\frac{1}{27}\int_1^4 16u^{-\frac{1}{2}}-8u^{\frac{1}{2}}+u^{\frac{3}{2}}\,du$$

Now you can apply the power rule to each term in succession.

 

[chisigma]

$\int_{0}^{1} \frac{y^2}{\sqrt{4-3y}}dy$

I know there is several ways to substitute this but I chose

$u=4-3y, du=3 dy, y^2=u-4$

The answer was $\frac{106}{405}$

But I couldn't get it.

A good alternative is to integrate by parts several times...

$\displaystyle \int x^{2}\ (4 - 3\ x)^{- \frac{1}{2}}\ d x = - \frac{2}{3}\ x^{2}\ (4 - 3\ x)^{\frac{1}{2}} + \frac{4}{3}\ \int x\ (4 - 3\ x)^{\frac{1}{2}}\ d x = $

$\displaystyle = - \frac{2}{3}\ x^{2}\ (4 - 3\ x)^{\frac{1}{2}} - \frac{8}{27}\ x\ (4 - 3\ x)^{\frac{3}{2}} + \frac{8}{27}\ \int (4 - 3\ x)^{\frac{3}{2}}\ dx = $

$\displaystyle = - \frac{2}{3}\ x^{2}\ (4 - 3\ x)^{\frac{1}{2}} - \frac{8}{27}\ x\ (4 - 3\ x)^{\frac{3}{2}} - \frac{16}{405}\ (4 - 3\ x)^{\frac{5}{2}} + c$

The computation of the definite integral is left to You...

Kind regards

$\chi$ $\sigma$

 

[karush]

$\frac{1}{27}\left[\frac{u}{4}
-\frac{-16{u}^{3/2}}{3}
+\frac{3u^{5 /3}}{5}\right]
+C$

Subst back in u and take $\left[1,4\right]$ and have answer

 

[soroban]

Hello, karush!

\displaystyle\int^1_0\frac{y^2}{\sqrt{4-3y}}\,dy

Whenever I have a square root of a linear function,
$\quad$ I prefer to let $u$ equal the entire radical.

$u \,=\,\sqrt{4-3y}\quad\Rightarrow\quad u^2 \,=\,4-3y $

$\quad y \:=\:\dfrac{4-u^2}{3} \quad\Rightarrow\quad dy \:=\:-\frac{2}{3}u\,du$

Substitute: $\displaystyle\:\int\frac{(\frac{4-u^2}{3})^2}{u}\left(-\tfrac{2}{3}u\,du\right) \;=\; -\tfrac{2}{27}\int(4-u^2)^2du $

This avoids those annoying fractional exponents.