What is area x+y=3,y^2=4x,y=0,1 ?

The answer to area x+y=3,y^2=4x,y=0,1 is 95/4

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area x+y=3,y^2=4x,y=0,1

area

x + y = 3, y^{2} = 4 x, y = 0, 1

\frac{95}{4}

Decimal Notation

23.75

Solution steps

Isolate

y

for

x + y = 3 : y = 3 - x

Isolate

y

for

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

f_{1} (x) = 3 - x

f_{2} (x) = 2 x

To find the intersection points solve

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

3 - x = 2 x : x = 1

3 - x = - 2 x : x = 9

3 - x = 0 : x = 3

3 - x = 1 : x = 2

2 x = - 2 x : x = 0

2 x = 0 : x = 0

2 x = 1 : x = \frac{1}{4}

- 2 x = 0 : x = 0

- 2 x = 1 :

No Solution for

x \in R

0 = 1 :

False

= \int_{0}^{\frac{1}{4}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{0}^{1} ∣ f_{2} (x) - f_{4} (x) ∣ d x + \int_{\frac{1}{4}}^{2} ∣ f_{3} (x) - f_{5} (x) ∣ d x + \int_{1}^{3} ∣ f_{1} (x) - f_{4} (x) ∣ d x + \int_{2}^{9} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{1}{4}} 2 x - (- 2 x) d x + \int_{0}^{1} 2 x - 0 d x + \int_{\frac{1}{4}}^{2} - 2 x - 1 d x + \int_{1}^{3} ∣ 3 - x - 0 ∣ d x + \int_{2}^{9} 3 - x - (- 2 x) d x

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

\int_{0}^{1} 2 x - 0 d x = \frac{4}{3}

\int_{\frac{1}{4}}^{2} - 2 x - 1 d x = \frac{19}{12} + \frac{8 2}{3}

\int_{1}^{3} ∣ 3 - x - 0 ∣ d x = 2

\int_{2}^{9} 3 - x - (- 2 x) d x = \frac{37}{2} - \frac{8 2}{3}

= \frac{1}{3} + \frac{4}{3} + \frac{19}{12} + \frac{8 2}{3} + 2 + \frac{37}{2} - \frac{8 2}{3}

Simplify

= \frac{95}{4}

\int (sin^{2} (x) + cos (x))^{2} d x

f (x) = x - 2 sin (x)

\int_{0}^{2} \frac{x}{x ^{4} + 1} d x

e^{- x^{2}}, 0

\frac{d}{d x} ((x^{4} + 1)^{\frac{1}{2}})