Answer:
Using either method, we obtain: [tex]t^\frac{3}{8}[/tex]
Step-by-step explanation:
a) By evaluating the integral:
[tex]\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du[/tex]
The integral itself can be evaluated by writing the root and exponent of the variable u as: [tex]\sqrt[8]{u^3} =u^{\frac{3}{8}[/tex]
Then, an antiderivative of this is: [tex]\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}[/tex]
which evaluated between the limits of integration gives:
[tex]\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}[/tex]
and now the derivative of this expression with respect to "t" is:
[tex]\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}[/tex]
b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:
"If f is continuous on [a,b] then
[tex]g(x)=\int\limits^x_a {f(t)} \, dt[/tex]
is continuous on [a,b], differentiable on (a,b) and [tex]g'(x)=f(x)[/tex]
Since this this function [tex]u^{\frac{3}{8}[/tex] is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:
[tex]\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}[/tex]
