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

The answer to area x=y^2+1,x=2y+1 is 4/3

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

area

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

Isolate

y

for

x = y^{2} + 1 : y = x - 1, y = - x - 1

Isolate

y

for

x = 2 y + 1 : y = \frac{x - 1}{2}

f_{1} (x) = x - 1

f_{2} (x) = - x - 1

To find the intersection points solve

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

x - 1 = - x - 1 : x = 1

x - 1 = \frac{x - 1}{2} : x = 5, x = 1

- x - 1 = \frac{x - 1}{2} : x = 1

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

= \int_{1}^{5} x - 1 - \frac{x - 1}{2} d x

\int_{1}^{5} x - 1 - \frac{x - 1}{2} d x = \frac{4}{3}

= \frac{4}{3}

\int \frac{26 x + 4}{( 5 x + 1 ) ( x - 1 )} d x

f (x) = \frac{1}{4} x^{4}

\frac{\partial}{\partial x} (e^{- x} sin (y))

\frac{\partial}{\partial x} (6 y^{8} x^{5} + 49 y x^{6})

y^{^{'}} = 8 + 2 x - 3 x^{2}