What is area-4x+x^2+10,12-5x,4x-6 ?

The answer to area-4x+x^2+10,12-5x,4x-6 is 9

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area-4x+x^2+10,12-5x,4x-6

area

- 4 x + x^{2} + 10, 12 - 5 x, 4 x - 6

9

Solution steps

To find the intersection points solve

f_{i} (x) = f_{j} (x)

- 4 x + x^{2} + 10 = 12 - 5 x : x = 1, x = - 2

- 4 x + x^{2} + 10 = 4 x - 6 : x = 4

12 - 5 x = 4 x - 6 : x = 2

= \int_{- 2}^{1} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{1}^{2} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{2}^{4} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 2}^{1} - 4 x + x^{2} + 10 - (12 - 5 x) d x + \int_{1}^{2} - 4 x + x^{2} + 10 - (12 - 5 x) d x + \int_{2}^{4} - 4 x + x^{2} + 10 - (4 x - 6) d x

\int_{- 2}^{1} - 4 x + x^{2} + 10 - (12 - 5 x) d x = \frac{9}{2}

\int_{1}^{2} - 4 x + x^{2} + 10 - (12 - 5 x) d x = \frac{11}{6}

\int_{2}^{4} - 4 x + x^{2} + 10 - (4 x - 6) d x = \frac{8}{3}

= \frac{9}{2} + \frac{11}{6} + \frac{8}{3}

Simplify

= 9

\int_{1}^{16} \frac{8}{x} d x

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

f (x) = e^{- 4 x}

\frac{d}{d x} (\frac{14 x + 8 x ^{2}}{( 7 + 8 x ) ^{2}})

x \to - 2 lim (\frac{x}{x ^{2} + 4})

What is area-4x+x^2+10,12-5x,4x-6 ?
The answer to area-4x+x^2+10,12-5x,4x-6 is 9