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| If $f$ is an even function then |
$\int_{-a}^{a}f(x)dx=0$ $\int_{-a}^{a}f(x)dx$=2$\int_{0}^{a}f(x)dx$ $f(x)=-f(-x)$ None of the above |
| $\int_{-a}^{a}f(x)dx$=2$\int_{0}^{a}f(x)dx$ |
| $\int_{-a}^{a}f(x)dx=\int_{-a}^{0}f(x)dx+\int_{0}^{a}f(x)dx$. In $\int_{-a}^{0}f(x)dx$ substitute $x=-z$, then the integral becomes $\int_{0}^{a}f(z)dz$ since $f$ is an even function(i.e $f(x)=f(-x)$). So the sum is 2$\int_{0}^{a}f(x)dx$. |
