What is area x=((y-3)^2)/5 ,x=13-y ?

The answer to area x=((y-3)^2)/5 ,x=13-y is 225/2

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area x=((y-3)^2)/5 ,x=13-y

area

x = \frac{( y - 3 ) ^{2}}{5}, x = 13 - y

\frac{225}{2}

Decimal Notation

112.5

Solution steps

Isolate

y

for

x = \frac{( y - 3 ) ^{2}}{5} : y = 5 x + 3, y = - 5 x + 3

Isolate

y

for

x = 13 - y : y = 13 - x

f_{1} (x) = 5 x + 3

f_{2} (x) = - 5 x + 3

To find the intersection points solve

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

5 x + 3 = - 5 x + 3 : x = 0

5 x + 3 = 13 - x : x = 5

- 5 x + 3 = 13 - x : x = 20

= \int_{0}^{5} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{5}^{20} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{0}^{5} 5 x + 3 - (- 5 x + 3) d x + \int_{5}^{20} - 5 x + 3 - (13 - x) d x

\int_{0}^{5} 5 x + 3 - (- 5 x + 3) d x = \frac{100}{3}

\int_{5}^{20} - 5 x + 3 - (13 - x) d x = \frac{475}{6}

= \frac{100}{3} + \frac{475}{6}

Simplify

= \frac{225}{2}

\frac{d}{d x} (\frac{3}{4} x^{\frac{1}{2}})

\int_{0}^{8} \frac{9}{10 x ^{2} + 11 x + 1} d x

\frac{d y}{d x} = \frac{x ^{2}}{6 e ^{y}}

f (x) = 4 x^{2} - 7 x

3 t sin (t)

What is area x=((y-3)^2)/5 ,x=13-y ?
The answer to area x=((y-3)^2)/5 ,x=13-y is 225/2