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

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

Popular >

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

area

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

\frac{9}{2}

Decimal Notation

4.5

Solution steps

Isolate

y

for

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

Isolate

y

for

x = y + 3 : y = x - 3

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 = x - 3 : x = 5

- x - 1 = x - 3 : x = 2

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

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

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

\int_{2}^{5} x - 1 - (x - 3) d x = \frac{19}{6}

= \frac{4}{3} + \frac{19}{6}

Simplify

= \frac{9}{2}

\frac{\partial}{\partial x} (6 x - 6 y + 5 x y)

\frac{d}{d x} (tan (sin (5 x)))

\int 6 x^{6} - x d x

\frac{\partial}{\partial x} (y^{2} + cos^{2} (x))

\int x 64 + x^{2} d x