Maharashtra State BoardHSC Arts 12th Board Exam
Advertisement Remove all ads

Evaluate the following : ∫01t51-t2⋅dt - Mathematics and Statistics

Advertisement Remove all ads
Advertisement Remove all ads
Sum

Evaluate the following : `int_0^1 t^5 sqrt(1 - t^2)*dt`

Advertisement Remove all ads

Solution

Let I = `int_0^1 t^5 sqrt(1 - t^2)*dt`

Put t = sin θ
∴ dt = cos θ dθ

When t = 1, θ = sin–11 = `pi/(2)`

When t = 0, θ = sin–10 = 0

∴ I = `int_0^(pi/2) sin^5 theta sqrt(1 - sin^2 theta)cos theta*d theta`

I = `int_0^(pi/2) sin^5 theta*cos theta* cos theta*d theta`

= `int_0^(pi/2) sin^5 theta(1 - sin^2 theta)*d theta`

= `int_0^(pi/2) (sin^5 theta - sin^7 theta)*d theta`

= `int_0^(pi/2) sin^5 theta*d theta - int_0^(pi/2) sin^7 thetad theta`.
Using Reduction formula, we get

I = `4/5*2/3 - 6/7*4/5*2/3`

= `(8)/(15)[1 - 6/7]`

= `(8)/(15) xx (1)/(7)`

= `(8)/(105)`.

Concept: Fundamental Theorem of Integral Calculus
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 4 Definite Integration
Miscellaneous Exercise 4 | Q 2.05 | Page 176
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×