[email protected]

+91-87964 74404

Grade 12Integral Calculus

Please solve the attached problem in integral calculus.

Question image for Please solve the attached problem in integral cal

P S Hamshika

8 Years agoGrade 12

1 Answer

Sami Ullah

8 Years ago

Its very easy just a few manipulations you’ll have to do.First I will just solve the integral and later apply the upper and lower limits.

$I=\int\frac{\sqrt{x}}{\sqrt{1-x^3}}.dx$

$Let$ $u=x^\frac{3}{2}$ $\Rightarrow$ $u^2=x^3$

$Then$ $du=\frac{3}{2}x^\frac{1}{2}.dx$

$\Rightarrow$ $\frac{2}{3}du=\sqrt{x}.dx$

$Now$ $substituting$ $values$ ,

$I=\int \frac{2}{3}\times \frac{1}{\sqrt{1-u^2}}.du$

$I= \frac{2}{3}\sin^{-1}(u)$

$I= \frac{2}{3}\sin^{-1}(x^\frac{3}{2})$

$Putting$ $values$ ,

$I= \frac{2}{3}[\sin^{-1}(1^\frac{3}{2})-\sin^{-1}(0)]$

$I= \frac{2}{3}[\frac{\pi }{2}-0]$

$I= \frac{\pi }{3}$

So the awnser is option (2).

The area of a square field is 640000cm².what is its area in hectares

integral calculus

1 Answer Available

Last Activity: 3 Years ago

Please solve this integration. I am facing difficulty in it
How to solve this. Please solve anyone. Humble request

integral calculus

1 Answer Available

Last Activity: 4 Years ago

Using laplace transform, solve d²y/dt²+ dy/dt = t² +2t given that y=4 and y’=-2 and t=0

integral calculus

1 Answer Available

Last Activity: 4 Years ago

If 4th term of an A.p is zero, show that the 8th term is double the 6th term

integral calculus

1 Answer Available

Last Activity: 4 Years ago

View all Questions