What is area f(x)=x^2-12,g(x)=x-6,[2sqrt(3),6] ?

The answer to area f(x)=x^2-12,g(x)=x-6,[2sqrt(3),6] is 24+4sqrt(3)

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area f(x)=x^2-12,g(x)=x-6,[2sqrt(3),6]

area

f (x) = x^{2} - 12, g (x) = x - 6, [23, 6]

24 + 43

Decimal

30.92820 \dots

Solution steps

Apply the area formula:

\int_{23}^{6} x^{2} - 12 - (x - 6) d x

Solve

\int_{23}^{6} x^{2} - 12 - (x - 6) d x : 24 + 43

The area is:

= 24 + 43

\int_{0}^{1} x - x d x

f (x) = 1 + cos (x), at x = \frac{3 π}{2}

\int x^{4} e^{- 2 x} d x

\int_{- 4}^{2} x^{2} + 2 d x

\frac{d}{d x} (\frac{A x + B}{C x + d})

What is area f(x)=x^2-12,g(x)=x-6,[2sqrt(3),6] ?
The answer to area f(x)=x^2-12,g(x)=x-6,[2sqrt(3),6] is 24+4sqrt(3)