What is area y=7-x,x=((y-4)^2)/4 ?

The answer to area y=7-x,x=((y-4)^2)/4 is 64/3

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area y=7-x,x=((y-4)^2)/4

area

y = 7 - x, x = \frac{( y - 4 ) ^{2}}{4}

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

Isolate

y

for

x = \frac{( y - 4 ) ^{2}}{4} : y = 2 x + 4, y = - 2 x + 4

f_{1} (x) = 7 - x

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

To find the intersection points solve

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

7 - x = 2 x + 4 : x = 1

7 - x = - 2 x + 4 : x = 9

2 x + 4 = - 2 x + 4 : x = 0

= \int_{0}^{1} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{1}^{9} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{1} 2 x + 4 - (- 2 x + 4) d x + \int_{1}^{9} 7 - x - (- 2 x + 4) d x

\int_{0}^{1} 2 x + 4 - (- 2 x + 4) d x = \frac{8}{3}

\int_{1}^{9} 7 - x - (- 2 x + 4) d x = \frac{56}{3}

= \frac{8}{3} + \frac{56}{3}

Simplify

= \frac{64}{3}

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

\frac{\partial}{\partial x} (6 x + 3 y)

\frac{d}{d x} (ln (x^{2} - 3))

y = 2 x^{4} - 16 x^{2} + 9, at x = - 4

y = \frac{2}{x}, (5, \frac{2}{5})

What is area y=7-x,x=((y-4)^2)/4 ?
The answer to area y=7-x,x=((y-4)^2)/4 is 64/3