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Popular >

area x=-y^2+4,x=-3y

Solution

area

x = - y^{2} + 4, x = - 3 y

Solution

\frac{125}{6}

+1

Solution steps

Isolate

y

for

x = - y^{2} + 4 : y = - x + 4, y = - - x + 4

Isolate

y

for

x = - 3 y : y = - \frac{x}{3}

f_{1} (x) = - x + 4

f_{2} (x) = - - x + 4

To find the intersection points solve

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

- x + 4 = - - x + 4 : x = 4

- x + 4 = - \frac{x}{3} : x = - 12

- - x + 4 = - \frac{x}{3} : x = 3

= \int_{- 12}^{3} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{3}^{4} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{- 12}^{3} - x + 4 - (- \frac{x}{3}) d x + \int_{3}^{4} - - x + 4 - - x + 4 d x

\int_{- 12}^{3} - x + 4 - (- \frac{x}{3}) d x = \frac{39}{2}

\int_{3}^{4} - - x + 4 - - x + 4 d x = \frac{4}{3}

= \frac{39}{2} + \frac{4}{3}

Simplify

= \frac{125}{6}

Popular Examples

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y^{3} d x + 2 (x^{3} - x y^{2}) d y = 0

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

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\frac{160 ( x + 1 )}{( x + 2 ) ^{2}}

y^{^{''}} - 16 y = 0

integral of (2x^2)/(2x^2+1)

\int \frac{2 x ^{2}}{2 x ^{2} + 1} d x

Frequently Asked Questions (FAQ)

What is area x=-y^2+4,x=-3y ?
The answer to area x=-y^2+4,x=-3y is 125/6