What is area x+1y=3,x+9=y^2 ?

The answer to area x+1y=3,x+9=y^2 is 343/6

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area x+1y=3,x+9=y^2

area

x + 1 y = 3, x + 9 = y^{2}

\frac{343}{6}

Decimal Notation

57.16666 \dots

Solution steps

Isolate

y

for

x + 1 y = 3 : y = 3 - x

Isolate

y

for

x + 9 = y^{2} : y = x + 9, y = - x + 9

f_{1} (x) = 3 - x

f_{2} (x) = x + 9

To find the intersection points solve

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

3 - x = x + 9 : x = 0

3 - x = - x + 9 : x = 7

x + 9 = - x + 9 : x = - 9

= \int_{- 9}^{0} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{0}^{7} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 9}^{0} x + 9 - (- x + 9) d x + \int_{0}^{7} 3 - x - (- x + 9) d x

\int_{- 9}^{0} x + 9 - (- x + 9) d x = 36

\int_{0}^{7} 3 - x - (- x + 9) d x = \frac{127}{6}

= 36 + \frac{127}{6}

Simplify

= \frac{343}{6}

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