What is area x+4y=25,x+7=y^2 ?

The answer to area x+4y=25,x+7=y^2 is 288

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area x+4y=25,x+7=y^2

area

x + 4 y = 25, x + 7 = y^{2}

288

Solution steps

Isolate

y

for

x + 4 y = 25 : y = \frac{25 - x}{4}

Isolate

y

for

x + 7 = y^{2} : y = x + 7, y = - x + 7

f_{1} (x) = \frac{25 - x}{4}

f_{2} (x) = x + 7

To find the intersection points solve

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

\frac{25 - x}{4} = x + 7 : x = 9

\frac{25 - x}{4} = - x + 7 : x = 57

x + 7 = - x + 7 : x = - 7

= \int_{- 7}^{9} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{9}^{57} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 7}^{9} x + 7 - (- x + 7) d x + \int_{9}^{57} \frac{25 - x}{4} - (- x + 7) d x

\int_{- 7}^{9} x + 7 - (- x + 7) d x = \frac{256}{3}

\int_{9}^{57} \frac{25 - x}{4} - (- x + 7) d x = \frac{608}{3}

= \frac{256}{3} + \frac{608}{3}

Simplify

= 288

\int \frac{1}{x ^{2}} 4 + \frac{1}{x} d x

- 2 x^{3} (x^{2} + 1)^{- 2}

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

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

\frac{\partial}{\partial x} (2 x y^{7} + x^{8} y^{7})