Question

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

Short answer

Hence, proved.

Step-by-step solution

Step 1. Given Information.

a and b are real numbers and a<b.

Step 2. Modulus property.

From the modulus property we know that,

$$\begin{gathered} -\left|f\left(x\right)\right|⩽f\left(x\right)⩽\left|f\left(x\right)\right|, \\ Applyingintegrationoverthisinequalityweget, \\ -{\int }_a^b\left|f\left(x\right)\right|dx⩽{\int }_a^{b}f\left(x\right)dx⩽{\int }_a^b\left|f\left(x\right)\right|dx. \\ Nowfromthisweneedonly, \\ {\int }_a^{b}f\left(x\right)dx⩽{\int }_a^b\left|f\left(x\right)\right|dx. \end{gathered}$$

Step 3. Proof.

As we all know,

$$\begin{gathered} \left|a\right|⩾a, \\ Thereforeleta={\int }_a^{b}f\left(x\right)dx. \\ Thismeans, \\ \left|{\int }_a^{b}f\left(x\right)dx\right|⩾{\int }_a^{b}f\left(x\right)dx. \\ Andalsofromstep2wehave, \\ {\int }_a^{b}f\left(x\right)dx⩽{\int }_a^b\left|f\left(x\right)\right|dx. \\ Combiningabovetworesultsweget, \\ {\int }_a^b\left|f\left(x\right)\right|dx⩾\left|{\int }_a^{b}f\left(x\right)dx\right|. \\ Hence,Proved. \end{gathered}$$