Question

Let g be a function whose domain on the closed interval [0, 8]. The graph of g is shown in the attachment. The graph is made up of straight line segments and a semicircle.

Let f(x)= ∫ [0, 2x-1] g(t)dt

Find f '(3).

Answer 1

the limit is from 0 to 2x-1 ?

Answer 2

we can use the fundamental theorem of calculus
$$
\Large{
\frac{d}{dx}\int_{a}^{x}g(t) ~dt = g(x)
}$$ small modification when the upper limit is a function of x
$$\Large{ \frac{d}{dx}\int_{a}^{u(x)}g(t) ~dt = g(u(x))\cdot u(x) '
}$$therefore we have
$$\Large{
f(x) = \int_{0}^{2x-1}g(t) ~dt
\\~\\f '(x) =\frac{d}{dx}\left( \int_{0}^{2x-1}g(t) ~dt\right)
\\~\\f '(x) =g(2x-1) \cdot (2x-1)' = g(2x-1) \cdot 2

\\~\\f '(3) =g(2\cdot 3 -1) \cdot 2 = g(5) \cdot 2


}$$
now use the curve

Answer 3

g(5)=1 so the answer is 2?

Answer 4

correct

Answer 5

Thanks a lot