Q. 79 - Calculus

Question

Prove that for all real numbers a and b with a < b, we have $\left|{\int }_a^{b}f\left(x\right)dx\right| ⩽{\int }_a^b\left|f\left(x\right)\right|dx.$

Step-by-Step Solution

Answer

Hence, proved.

Step 1. Given Information.

a and b are real numbers and a<b.

Step 2. Modulus property.

From the modulus property we know that,

$$\begin{gathered} -\left|f\left(x\right)\right| ⩽f\left(x\right) ⩽\left|f\left(x\right)\right|, \\ Applying integration over this inequality we get, \\ -{\int }_a^b\left|f\left(x\right)\right|dx ⩽{\int }_a^{b}f\left(x\right)dx⩽{\int }_a^b\left|f\left(x\right)\right|dx. \\ Now from this we need only, \\ {\int }_a^{b}f\left(x\right)dx ⩽{\int }_a^b\left|f\left(x\right)\right|dx. \end{gathered}$$

Step 3. Proof.

As we all know,

$$\begin{gathered} \left|a\right|⩾a, \\ Therefore let a = {\int }_a^{b}f\left(x\right)dx. \\ This means, \\ \left|{\int }_a^{b}f\left(x\right)dx\right| ⩾ {\int }_a^{b}f\left(x\right)dx. \\ And also from step 2 we have, \\ {\int }_a^{b}f\left(x\right)dx ⩽{\int }_a^b\left|f\left(x\right)\right|dx. \\ Combining above two results we get, \\ {\int }_a^b\left|f\left(x\right)\right|dx ⩾ \left|{\int }_a^{b}f\left(x\right)dx\right|. \\ Hence, Proved. \end{gathered}$$