What is area y^2=x-2,y=x-4 ?

The answer to area y^2=x-2,y=x-4 is 9/2

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area y^2=x-2,y=x-4

area

y^{2} = x - 2, y = x - 4

\frac{9}{2}

Decimal Notation

4.5

Solution steps

Isolate

y

for

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

f_{1} (x) = x - 2

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

To find the intersection points solve

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

x - 2 = - x - 2 : x = 2

x - 2 = x - 4 : x = 6

- x - 2 = x - 4 : x = 3

= \int_{2}^{3} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{3}^{6} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{2}^{3} x - 2 - (- x - 2) d x + \int_{3}^{6} x - 2 - (x - 4) d x

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

\int_{3}^{6} x - 2 - (x - 4) d x = \frac{19}{6}

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

Simplify

= \frac{9}{2}

\int (sec (2 x) + tan (2 x)) d x

f (x) = \frac{2 t + 1}{t + 3}

y = 2 x^{2} + 3, (0, 3)

\int \frac{x ^{2} + 1}{1 - x ^{2}} d x

x \to \frac{5 π}{6} lim (sin (x))