Question

Integrate it ✌✌✌✌✌✌​

Answer 1

Answer:

0 is the answer.

Plz refer to the attachment for required solution ✨✨✨

Answer 2

Answer:

0

Step-by-step explanation:

You don't need to integrate at all, it can be solved by using properties of definite integrals alone!!

Here is how:-

using kings rule :-

$\int\limits^a_b {x} \, dx =\int\limits^a_b {(a+b-x)} \, dx$

Using this you get,

I = $\int\limits^2_a {|xcos(\pi x )|} \, dx$-----(1)

I = $\int\limits^2_a {|(2-2-x)cos\pi(2-2- x )|} \, dx$

I =$\int\limits^2_a {|-xcos(\pi x )|} \, dx$-----(2)

Adding (1) and (2), we get

2I=$\int\limits^2_a {|xcos(\pi x )|} \, dx$+$\int\limits^2_a {|-xcos(\pi x )|} \, dx$

Since |y|+|-y|=0 for all y,

2I=0

∴I=0

Hope this answer helped you :)