Question

Evaluate the Integrals

Answer 1

Solution :

Let, $\large{\mathsf{30-x^{\frac{3}{2}}=z}}$

Taking differentials, we get

$\large{\mathsf{-\frac{3}{2}x^{\frac{1}{2}}dx=dz}}$

↬ $\large{\mathsf{\sqrt{x}dx=-\frac{2}{3}dz}}$

As x tends to 4, z tends to 22

As x tends to 9, z tends to 3

Then, $\large{\mathsf{\int \frac{\sqrt{x}}{(30-x^{\frac{3}{2}})^{2}}dx}}$

$\large{\mathsf{=\int \frac{-\frac{2}{3}dz}{z^{2}}}}$

$\large{\mathsf{=-\frac{2}{3} \int \frac{dz}{z^{2}}}}$

$\large{\mathsf{=-\frac{2}{3}(-\frac{1}{z})}}$

$\large{\mathsf{=\frac{2}{3z}}}$

Thus, $\large{\mathsf{\int_{4}^{9} \frac{\sqrt{x}}{(30-x^{\frac{3}{2})^{2}}}dx}}$

$\large{=\mathsf{[\frac{2}{3z}]_{22}^{3}}}$

$\large{\mathsf{=\frac{2}{9}-\frac{2}{66}}}$

$\large{\mathsf{=\frac{2}{9}-\frac{1}{33}}}$

$\large{=\mathsf{\frac{22-3}{99}}}$

$\large{\mathsf{=\frac{19}{99}}}$

$\to \boxed{\large{\mathsf{\int_{4}^{9} \frac{\sqrt{x}}{(30-x^{\frac{3}{2}})^{2}}dx=\frac{19}{99}}}}$

Answer 2

Answer: 19/99

Step-by-step explanation:

Let,

$I=\int\limits^9_4{\frac{\sqrt{x}}{(30-x^{\frac{3}{2}})^2}}\,dx$

Let,

$30-x^{\frac{3}{2}}=t\\\;\\-\frac{3}{2}x^{\frac{1}{2}}=\frac{dt}{dx}\\\;\\\sqrt{x}\;dx=-\frac{2dt}{3}$

Also,

When, x = 9 , t = 3

x = 4 , t = 22,

Now,

$I=\int\limits^3_{22}{\frac{-\frac{2dt}{3}}{t^2}}\\\;\\I=-\frac{2}{3}\int\limits^3_{22}{\frac{dt}{t^2}}\\\;\\I=\frac{2}{3}\int\limits^{22}_3{\frac{dt}{t^2}}\\\;\\I=\frac{2}{3}[-\frac{1}{t}]\limits^{22}_3\\\;\\I=-\frac{2}{3}[\frac{1}{t}]\limits^{22}_3\\\;\\I=-\frac{2}{3}[\frac{1}{22}-\frac{1}{3}]\\\;\\I=\frac{2}{3}[\frac{1}{3}-\frac{1}{22}]\\\;\\I=\frac{2}{3}[\frac{22-3}{66}]\\\;\\I=\frac{2}{3}\times\frac{19}{66}\\\;\\I=\frac{19}{99}$