Question

The Fundamental Theorem of Calculus does not apply to the integral ${\int }_0^5{\left(x-3\right)}^{4/3}dx$; why not? What would the Fundamental Theorem of Calculus calculate for this integral if we applied it anyway? Would it give the correct answer?

Short answer

The Fundamental Theorem of Calculus is not applied to the given integral because It give wrong answer because the function is not continuous at $x=3$inside the integral.

Step-by-step solution

Step 1. Given information.  

We are given,

${\int }_0^5{\left(x-3\right)}^{4/3}dx$

Step 2. Calculating the integral using Fundamental Theorem of Calculus

The Fundamental Theorem says, if f is continuous on $[a,b]$ and F is any antiderivative of f, then

role="math" ${\int }_a^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right)$

So,

The integrand $(x-3{)}^{-\frac{4}{3}}$is not continuous at $x=3$.

So, the Fundamental theorem of calculus does not apply in the above integral because $(x-3{)}^{-\frac{4}{3}}$is not continuous at $x=3$, inside the interval $[0,5]$.

Applying it the interval give the incorrect answer $\frac{-3}{2^{\frac{1}{3}}}+\frac{3}{(-3{)}^{\frac{1}{3}}}$, when the correct answer is that the improper integral diverges.