area x=y^2-4y,x=2y-y^2 — step-by-step solution

⚠ 4y is read as the product 4*y: for a power, write 4^y

⚠ 2y is read as the product 2*y: for a power, write 2^y

x

is already isolated in

x = y^{2} - 4 y

x

is already isolated in

x = 2 y - y^{2}

At the intersections,

y^{2} - 4 y

2 y - y^{2}

. Subtract the second expression from the first:

(y^{2} - 4 y) - (2 y - y^{2})

= 0, giving

2 y^{2} - 6 y = 0

⇥1.1) Let's solve in

ℝ

2 y^{2} - 6 y = 0

⇥You can factorize the expression:

2 y^{2} - 6 y

into:

y \cdot (2 y - 6)

⇥Solution found:

y = 0

⇥1.2) Let's solve in

ℝ

2 y - 6 = 0

⇥

2 y = 6

⇥

y = \frac{6}{2}

⇥Solution obtained:

y = 3

The area between

y^{2} - 4 y

and

2 y - y^{2}

[0, 3]

is the integral of the absolute value of their difference

No interior intersection: the difference keeps its sign on the interval

[0, 3]

2 y - y^{2}

is above

y^{2} - 4 y

: integrate

- 2 y^{2} + 6 y

⇥1.3) Let's compute the integral:

\int(- 2 y^{2} + 6 y)dy

⇥

- \int(2 y^{2})dy + \int(6 y)dy

⇥ℹ

- \int(2 y^{2})dy

- 2 \cdot \frac{y^{3}}{3}

‖

\int(6 y)dy

6 \cdot \frac{y^{2}}{2}

⇥

- 2 \cdot \frac{y^{3}}{3} + 6 \cdot \frac{y^{2}}{2}

⇥ℹ

- 2 \cdot \frac{y^{3}}{3}

- \frac{2 y^{3}}{3}

‖

6 \cdot \frac{y^{2}}{2}

\frac{6 y^{2}}{2}

‖

\frac{6 y^{2}}{2}

3 y^{2}

▷

- \frac{2 y^{3}}{3} + 3 y^{2}

⇥1.4) Let's evaluate for the interval

[0, 3]

(- \frac{2 \cdot 3^{3}}{3} + 3 \cdot 3^{2}) - (- \frac{2 \cdot 0^{3}}{3} + 3 \cdot 0^{2})

⇥ℹ

3^{3}

27

‖

3^{2}

9

‖

- (- \frac{2 \cdot 0^{3}}{3} + 3 \cdot 0^{2})

\frac{2 \cdot 0^{3}}{3} - 3 \cdot 0^{2}

⇥

- \frac{2 \cdot 27}{3}

+ 3 \cdot 9

+ \frac{2 \cdot 0^{3}}{3}

- 3 \cdot 0^{2}

⇥ℹ

2 \cdot 27

54

‖

3 \cdot 9

27

‖

2 \cdot 0

0

‖

-3 \cdot 0

0

⇥

- \frac{54}{3}

+ 27

+ 0

+ 0

⇥

- 18 + 27

⇥

9

The area on

[0, 3]

9

▶Area:

9