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Popular >

area-2x^2+8,3x+1,0

Solution

area

- 2 x^{2} + 8, 3 x + 1, 0

Solution

\frac{287}{24}

+1

Decimal

11.95833 \dots

Solution steps

To find the intersection points solve

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

- 2 x^{2} + 8 = 3 x + 1 : x = - \frac{3 + 65}{4}, x = \frac{65 - 3}{4}

- 2 x^{2} + 8 = 0 : x = 2, x = - 2

3 x + 1 = 0 : x = - \frac{1}{3}

= \int_{- \frac{3 + 65}{4}}^{- 2} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{- 2}^{- \frac{1}{3}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{- \frac{1}{3}}^{\frac{65 - 3}{4}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{65 - 3}{4}}^{2} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- \frac{3 + 65}{4}}^{- 2} - 2 x^{2} + 8 - (3 x + 1) d x + \int_{- 2}^{- \frac{1}{3}} ∣ 3 x + 1 - 0 ∣ d x + \int_{- \frac{1}{3}}^{\frac{65 - 3}{4}} ∣ 3 x + 1 - 0 ∣ d x + \int_{\frac{65 - 3}{4}}^{2} - 2 x^{2} + 8 - 0 d x

\int_{- \frac{3 + 65}{4}}^{- 2} - 2 x^{2} + 8 - (3 x + 1) d x = - \frac{44}{3} + \frac{279 + 65 65}{48}

\int_{- 2}^{- \frac{1}{3}} ∣ 3 x + 1 - 0 ∣ d x = \frac{25}{6}

\int_{- \frac{1}{3}}^{\frac{65 - 3}{4}} ∣ 3 x + 1 - 0 ∣ d x = \frac{1}{6} + \frac{99 - 5 65}{16}

\int_{\frac{65 - 3}{4}}^{2} - 2 x^{2} + 8 - 0 d x = \frac{247 - 25 65}{24}

= - \frac{44}{3} + \frac{279 + 65 65}{48} + \frac{25}{6} + \frac{1}{6} + \frac{99 - 5 65}{16} + \frac{247 - 25 65}{24}

Simplify

= \frac{287}{24}

Graph

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Frequently Asked Questions (FAQ)

What is area-2x^2+8,3x+1,0 ?
The answer to area-2x^2+8,3x+1,0 is 287/24