What is area x=2+sqrt(y+3),x=2-sqrt(y+3),x=y-1 ?

The answer to area x=2+sqrt(y+3),x=2-sqrt(y+3),x=y-1 is 125/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area x=2+sqrt(y+3),x=2-sqrt(y+3),x=y-1

area

x = 2 + y + 3, x = 2 - y + 3, x = y - 1

\frac{125}{6}

Solution steps

Isolate

y

for

x = 2 + y + 3 : y = x^{2} - 4 x + 1 {x \geq 2}

Isolate

y

for

x = 2 - y + 3 : y = x^{2} - 4 x + 1 {x \leq 2}

Isolate

y

for

x = y - 1 : y = x + 1

Apply the area formula:

\int_{0}^{5} x^{2} - 4 x + 1 - (x + 1) d x

Solve

\int_{0}^{5} x^{2} - 4 x + 1 - (x + 1) d x : \frac{125}{6}

The area is:

= \frac{125}{6}

\frac{d}{d x} ((- 3 x^{5} + 1)^{3})

\int \frac{x}{( sin ( x ^{2} ) ) ^{2}} d x

\frac{\partial}{\partial x} (y + x cos (t))

\frac{\partial}{\partial x} (tan (2 x + y))

y = \frac{x - 6}{x + 6}

What is area x=2+sqrt(y+3),x=2-sqrt(y+3),x=y-1 ?
The answer to area x=2+sqrt(y+3),x=2-sqrt(y+3),x=y-1 is 125/6