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

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

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

area

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

\frac{225}{2}

Decimal

112.5

Solution steps

Isolate

y

for

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

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}

\int + \frac{13}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

x \to 1 lim (\frac{1}{x})

n \to \infty lim (tanh (n))

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

\int 4 e^{- t} t d t

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