Question

Use Part ofthe Fundamental Theorem of Calculus to find che derivative of the function

Answer 1

The fundamental theorem of calculus states that if f is a continuous real-valued function defined in a closed interval [a,b], and F is the function defined on [a,x] by

$F(x)=\int ^x_af(t)dt$

Then, F is uniformly continuous on [a,b] and differentiable on (a,b), and

$F^{\prime}(x)=f(x);x\in(a,b)$

In our case,

$f(t)=\frac{1}{3}(t^2+\sqrt[]{t})$

is continuous in the interval (0,infinite); so, it is continuous in [2,infinite)

Therefore,

$\Rightarrow F(x)=\int ^x_2\frac{1}{3}(t^2+\sqrt[]{t})dt$

Thus,

$\Rightarrow F^{\prime}(x)=\frac{1}{3}(x^2+\sqrt[]{x})$

The answer is F'(x)=(x^2+sqrt(x))/3