$\int_{0}^{3} \sin ^{39}(x) d x\pi$ .

Question

$\int_{0}^{3} \sin ^{39}(x) d x>\pi$ .

Quizplus · Accepted Answer

The integral of $\sin^{39}(x)$ from 0 to 3 is a definite integral of a bounded function (since $\sin(x)$ is bounded between -1 and 1) over a finite interval. The value of this integral cannot exceed the length of the interval multiplied by the maximum value of the function, which in this case is less than 3 (the length of the interval [0,3]). Since $\pi$ is approximately 3.14, the integral's value cannot be greater than $\pi$.

∫03sin⁡39(x)dx>π\int_{0}^{3} \sin ^{39}(x) d x>\pi∫03​sin39(x)dx>π

$\int_{0}^{3} \sin ^{39}(x) d x>\pi$