What is area 2y=4sqrt(x),2y+x=5,y=5 ?

The answer to area 2y=4sqrt(x),2y+x=5,y=5 is 63/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area 2y=4sqrt(x),2y+x=5,y=5

area

2 y = 4 x, 2 y + x = 5, y = 5

\frac{63}{4}

Decimal Notation

15.75

Solution steps

Isolate

y

for

2 y = 4 x : y = 2 x

Isolate

y

for

2 y + x = 5 : y = \frac{5 - x}{2}

f_{1} (x) = 2 x

f_{2} (x) = \frac{5 - x}{2}

To find the intersection points solve

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

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

2 x = 5 : x = \frac{25}{4}

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

= \int_{- 5}^{1} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{1}^{\frac{25}{4}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 5}^{1} \frac{5 - x}{2} - 5 d x + \int_{1}^{\frac{25}{4}} 2 x - 5 d x

\int_{- 5}^{1} \frac{5 - x}{2} - 5 d x = 9

\int_{1}^{\frac{25}{4}} 2 x - 5 d x = \frac{27}{4}

= 9 + \frac{27}{4}

Simplify

= \frac{63}{4}

x \to 4 lim (3 x^{3} - 5 x^{2} + x - 5)

\frac{d}{d θ} (cos (θ^{2}))

\frac{9}{3 s - 1}

\int x^{1} d x

What is area 2y=4sqrt(x),2y+x=5,y=5 ?
The answer to area 2y=4sqrt(x),2y+x=5,y=5 is 63/4