What is area y=-x^2+2x+5,y=(x^2)/2-4x+5 ?

The answer to area y=-x^2+2x+5,y=(x^2)/2-4x+5 is 16

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area y=-x^2+2x+5,y=(x^2)/2-4x+5

area

y = - x^{2} + 2 x + 5, y = \frac{x ^{2}}{2} - 4 x + 5

16

Solution steps

Apply the area formula:

\int_{0}^{4} - x^{2} + 2 x + 5 - (\frac{x ^{2}}{2} - 4 x + 5) d x

Solve

\int_{0}^{4} - x^{2} + 2 x + 5 - (\frac{x ^{2}}{2} - 4 x + 5) d x : 16

The area is:

= 16

x \to 0 lim (ln (1 + \frac{1}{x}))

\int e^{- 7 t} d t

16 y^{^{''}} + 2 y = 0

f (x) = - 9 x^{2}

\frac{d}{d x} (x^{2} - 2 x^{3})

What is area y=-x^2+2x+5,y=(x^2)/2-4x+5 ?
The answer to area y=-x^2+2x+5,y=(x^2)/2-4x+5 is 16