Q.

If -0.150.15|100x2-1|dx=k3000, then k is equal to ___________.      [2023]


Ans.

(575)

Let I=-0.150.15|100x2-1|dx

=200.15|100x2-1|dx

(-aaf(x)dx={20af(x)dx, if f(-x)=f(x) i.e. f(x) is even0,if f(-x)=-f(x) i.e. f(x) is odd)

=2{00.1-(100x2-1)dx+0.10.15(100x2-1)dx}

=2{[x-100x33]00.1+[100x33-x]0.10.15}=5753000

k=575