0

Guest

Courses New

Solve ∫|sin2pix| dx Limit is from 0 to 1.

Solve
∫|sin2pix| dx
Limit is from 0 to 1.

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

I = \int_{0}^{1}|sin2\pi x|dx
Period of this function is ½.
I = 2\int_{0}^{1/2}|sin2\pi x|dx
And it is +ve b/w 0 and ½.
I = 2\int_{0}^{1/2}(sin2\pi x)dx
I = 2(\frac{-cos2\pi x}{2\pi })_{0}^{1/2}
I = (\frac{-cos2\pi x}{\pi })_{0}^{1/2}
I = \frac{2}{\pi }

Think You Can Provide A Better Answer ?

Other Related Questions on Integral Calculus

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More