G.4.d.i) Experienced integral watchers know how to break the code of the integra

Question

G.4.d.i)

Experienced integral watchers know how to break the code of the integral. The first step toward breaking the code of the integral is to learn how to calculate
f'[x]
when f[x] is given by
      f[x] = Integrate[g[t], {t, a, x}]
for some other function g[x].
Go with the specific case of
      f[x] = Integrate[g[t], {t, a, x}]
withg[x] = Cos[3 x] + 1/2
and a = 0.
Here's a plot of g[x] for axb = 4:

And here comes a plot of
      f[x] = Integrate[g[t], {t, a, x}]for a x b:

Here they are together:

Describe what you see, paying special attention to what
      f[x] = Integrate[g[t], {t, a, x}](thick)
is doing when g[x] (thin) is positive and to what
      f[x] = Integrate[g[t], {t, a, x}]
is doing when g[x] is negative.
What clue does this give you about the relationship between
f'[x] and g[x]?

a = 0; Clear[f, g, x, t]; g[x]- Cos[3 x]1/2; gplot Plot[g(x], {x, a, b), Plotstyle-) {{Red, Thickness(0.01))), AxesLabel-> {"x", "glxr1 (Red, Thicknesste.on])m g[x] 1.5 1.0 0.5 -0.5

Explanation / Answer

g (x) is a curve of Cos (3x) and +1/2 so its uppre limit and lower limit is constant.

And it is similar to cosx curve.

But f'(x) is a curve of -3Sin (3x) and x/2 and a constant , here constant is zero (0).

Un this above curve when g (x) is -ve, f'(x) is +ve value and as we increase the value of x the value of f'(x) is going high.

When g (x) is +ve the value of f'(x) is +ve and very high.

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