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Popular Calculus >

area 3x+y^2=13,x=4y

Solution

area

3 x + y^{2} = 13, x = 4 y

Solution

\frac{1372}{9}

+1

Decimal

152.44444 \dots

Solution steps

Isolate

y

for

3 x + y^{2} = 13 : y = 13 - 3 x, y = - 13 - 3 x

Isolate

y

for

x = 4 y : y = \frac{x}{4}

f_{1} (x) = 13 - 3 x

f_{2} (x) = - 13 - 3 x

To find the intersection points solve

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

13 - 3 x = - 13 - 3 x : x = \frac{13}{3}

13 - 3 x = \frac{x}{4} : x = 4

- 13 - 3 x = \frac{x}{4} : x = - 52

= \int_{- 52}^{4} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{4}^{\frac{13}{3}} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 52}^{4} - 13 - 3 x - \frac{x}{4} d x + \int_{4}^{\frac{13}{3}} 13 - 3 x - (- 13 - 3 x) d x

\int_{- 52}^{4} - 13 - 3 x - \frac{x}{4} d x = 152

\int_{4}^{\frac{13}{3}} 13 - 3 x - (- 13 - 3 x) d x = \frac{4}{9}

= 152 + \frac{4}{9}

Simplify

= \frac{1372}{9}

Graph

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Frequently Asked Questions (FAQ)

What is area 3x+y^2=13,x=4y ?
The answer to area 3x+y^2=13,x=4y is 1372/9