Question

The value of the integral ${\int }_{\pi }^{0}cos(\pi -x)$ will be equal to the value of which of the following integral.

• A. ${\int }_0^{\pi }cos\left(x\right)dx$
• B. $-{\int }_0^{\pi }cos\left(x\right)dx$
• C. ${\int }_0^{\pi }sin\left(x\right)dx$
• D. $-{\int }_0^{\pi }sin\left(x\right)dx$

Answer 1

The correct option is A ${\int }_0^{\pi }cos\left(x\right)dx$

This is only a two step problem if we understand the fact on interchanging the lower and upper limits adds a negative sign to the expression.
i.e., ${\int }_a^{b}f\left(x\right).dx=-{\int }_b^{a}f\left(x\right).dxSohere{\int }_{\pi }^{0}cos(\pi -x)dx={\int }_{\pi }^0-cos\left(x\right)dx={\int }_0^{\pi }cos\left(x\right)dx$