Question

For each pair of definite integrals in exercise 13-18 decide which if either is larger without computing the integrals

$\pi {\int }_0^1{e}^{2x}dxand\pi {\int }_0^1(e^{2x}-1)dx$

Short answer

The integral$\pi {\int }_0^1{e}^{2x}dx$is bigger

Step-by-step solution

Given information

We are given two integrals

$\pi {\int }_0^1{e}^{2x}dxand\pi {\int }_0^1(e^{2x}-1)dx$

Explaination

On integrating the second integral we get an extra term of x. And after substituting the limits the value of second integral will be 1 less than that of the first integral. Hence the first integral is bigger.