Question

If $f\left(x\right)$is defined at$x=a$, then${\int }_a^{a}f\left(x\right)dx=0$. Explain why this makes sense in terms of area.

Short answer

The width is zero so the area is zero.

Therefore the value is zero.

Step-by-step solution

Step 1. Given information

An expression is given as${\int }_a^{a}f\left(x\right)dx=0$

Step 2. Drawing area

The integral is defined as

${\int }_a^{b}f\left(x\right)dx=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k^{*}\right)\Delta x$

where $\Delta x=\frac{b-a}{n},x_k=a+k\Delta x\&x_k^{*}=x_k$

So, the function fon the interval $[a,a]$ and any nrectangles,

$\Delta x=\frac{a-a}{n}=0$

Since the width is zero therefore the area is zero,

${\int }_a^{a}f\left(x\right)dx=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k^{*}\right)\Delta x=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k^{*}\right)\left(0\right)=0$