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$\int_{- \pi }^{\pi} t^{2} \sin t \ dt$
asked in Calculus | 67 views
Use the rule $\int_{-a}^{a} f(x)$ = {0, if f(-x)=-f(x); 2$\int_{0}^{a} f(x)$ if f(-x)=f(x)}

See the basic property :

-a∫ a  f(x)  dx =   0 , f(x) is odd

=  2 0a f(x) dx

So here :

The function f(x)  =  x2 sinx  is odd as we know product of odd function and even function is odd function only..

Therefore the given integral value =  0.