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Let 
. Then the definite integral can be split up at 
so that
Now take the derivative. By the fundamental theorem of calculus, you have
![\displaystyle\frac{\mathrm dF}{\mathrm dx}=\frac{\mathrm d}{\mathrm dx}\left[\int_{t=c}^{t=5x}\frac{\mathrm dt}t-\int_{t=c}^{t=2x}\frac{\mathrm dt}t\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B%5Cmathrm%20dF%7D%7B%5Cmathrm%20dx%7D%3D%5Cfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cint_%7Bt%3Dc%7D%5E%7Bt%3D5x%7D%5Cfrac%7B%5Cmathrm%20dt%7Dt-%5Cint_%7Bt%3Dc%7D%5E%7Bt%3D2x%7D%5Cfrac%7B%5Cmathrm%20dt%7Dt%5Cright%5D)
![=\displaystyle\frac1{5x}\dfrac{\mathrm d}{\mathrm dx}[5x]-\frac1{2x}\dfrac{\mathrm d}{\mathrm dx}[2x]](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cfrac1%7B5x%7D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5B5x%5D-%5Cfrac1%7B2x%7D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5B2x%5D)
Then integrating with respect to
, we recover 
and find that
where
is an arbitrary constant.
Now take the derivative. By the fundamental theorem of calculus, you have
Then integrating with respect to
where
Answer:
x =
Step-by-step explanation:
Given a quadratic in standard form, ax² + bx + c : a ≠ 0
Then the axis of symmetry is a vertical line with equation x = h
where h is the x- coordinate of the vertex.
The x- coordinate of the vertex is
= -
f(x) = 5x² - 3x + 5 ← is in standard form
with a = 5 and b = - 3, thus
= -
=
Thus the equation of the axis of symmetry is x =