What is area y=2x^2,y=8x^2,5x+y=3 ?

The answer to area y=2x^2,y=8x^2,5x+y=3 is 4157/384

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area y=2x^2,y=8x^2,5x+y=3

area

y = 2 x^{2}, y = 8 x^{2}, 5 x + y = 3

\frac{4157}{384}

Decimal

10.82552 \dots

Solution steps

Isolate

y

for

5 x + y = 3 : y = 3 - 5 x

f_{1} (x) = 2 x^{2}

f_{2} (x) = 8 x^{2}

To find the intersection points solve

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

2 x^{2} = 8 x^{2} : x = 0

2 x^{2} = 3 - 5 x : x = \frac{1}{2}, x = - 3

8 x^{2} = 3 - 5 x : x = \frac{3}{8}, x = - 1

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

= \int_{- 3}^{- 1} 2 x^{2} - (3 - 5 x) d x + \int_{- 1}^{0} 8 x^{2} - 2 x^{2} d x + \int_{0}^{\frac{3}{8}} 2 x^{2} - 8 x^{2} d x + \int_{\frac{3}{8}}^{\frac{1}{2}} 2 x^{2} - (3 - 5 x) d x

\int_{- 3}^{- 1} 2 x^{2} - (3 - 5 x) d x = \frac{26}{3}

\int_{- 1}^{0} 8 x^{2} - 2 x^{2} d x = 2

\int_{0}^{\frac{3}{8}} 2 x^{2} - 8 x^{2} d x = \frac{27}{256}

\int_{\frac{3}{8}}^{\frac{1}{2}} 2 x^{2} - (3 - 5 x) d x = \frac{41}{768}

= \frac{26}{3} + 2 + \frac{27}{256} + \frac{41}{768}

Simplify

= \frac{4157}{384}

\frac{d}{d x} (2 e^{x} + x e^{x})

f (x) = 8 + 5 x^{2} - 2 x^{3}, (2, 12)

\int_{7}^{9} \frac{7}{( x - 8 ) ^{10}} d x

\int \frac{( x ^{2} - x + 42 )}{x ^{3} + 7 x} d x

\frac{d}{d x} (- (x + 1) (x - 1)^{2})

What is area y=2x^2,y=8x^2,5x+y=3 ?
The answer to area y=2x^2,y=8x^2,5x+y=3 is 4157/384