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

1) Isolate y in

y = x^{2}

y = x^{2}

Isolate y in

x = \frac{y}{2}

y = 2 x

Find the intersections:

x^{2} - 2 x = 0

⇥You can factorize the expression:

x^{2} - 2 x

into:

x \cdot (x - 2)

⇥Solution obtained:

x = 0

⇥1.1) Let's solve in

ℝ

x - 2 = 0

⇥Solution obtained:

x = 2

The area between

x^{2}

and

2 x

[0, 2]

is the integral of the absolute value of their difference

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

[0, 2]

2 x

is above

x^{2}

: integrate

- x^{2} + 2 x

⇥1.2) Let's integrate:

\int(- x^{2} + 2 x)dx

⇥

- \int(x^{2})dx + \int(2 x)dx

⇥ℹ

- \int(x^{2})dx

- \frac{x^{3}}{3}

‖

\int(2 x)dx

2 \cdot \frac{x^{2}}{2}

⇥

- \frac{x^{3}}{3} + 2 \cdot \frac{x^{2}}{2}

⇥ℹ

2 \cdot \frac{x^{2}}{2}

\frac{2 x^{2}}{2}

‖

\frac{2 x^{2}}{2}

x^{2}

⇥

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

⇥1.3) Let's evaluate over the interval

[0, 2]

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

⇥ℹ

2^{3}

8

‖

2^{2}

4

‖

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

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

⇥

- \frac{8}{3}

+ 4

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

- 0^{2}

⇥

\frac{- 8 + 4 \cdot 3}{3} + 0

⇥

\frac{- 8 + 12}{3}

⇥

\frac{4}{3}

The area on

[0, 2]

\frac{4}{3}

▶Area:

\frac{4}{3}