What is area 10-7x,y=0,x^2-20 ?

The answer to area 10-7x,y=0,x^2-20 is 11821/42

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area 10-7x,y=0,x^2-20

area

10 - 7 x, y = 0, x^{2} - 20

\frac{11821}{42}

Decimal

281.45238 \dots

Solution steps

To find the intersection points solve

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

0 = 10 - 7 x : x = \frac{10}{7}

0 = x^{2} - 20 : x = 25, x = - 25

10 - 7 x = x^{2} - 20 : x = 3, x = - 10

= \int_{- 10}^{- 25} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{- 25}^{\frac{10}{7}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{\frac{10}{7}}^{3} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{3}^{25} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 10}^{- 25} 10 - 7 x - (x^{2} - 20) d x + \int_{- 25}^{\frac{10}{7}} ∣ 0 - (10 - 7 x) ∣ d x + \int_{\frac{10}{7}}^{3} ∣ 0 - (10 - 7 x) ∣ d x + \int_{3}^{25} 0 - (x^{2} - 20) d x

\int_{- 10}^{- 25} 10 - 7 x - (x^{2} - 20) d x = - \frac{1000 - 40 5}{3} + 580 - 605

\int_{- 25}^{\frac{10}{7}} ∣ 0 - (10 - 7 x) ∣ d x = 205 + \frac{540}{7}

\int_{\frac{10}{7}}^{3} ∣ 0 - (10 - 7 x) ∣ d x = \frac{121}{14}

\int_{3}^{25} 0 - (x^{2} - 20) d x = - \frac{40 5}{3} - 51 + 405

= - \frac{1000 - 40 5}{3} + 580 - 605 + 205 + \frac{540}{7} + \frac{121}{14} - \frac{40 5}{3} - 51 + 405

Simplify

= \frac{11821}{42}