area 2x^2,2x+6,-1,0 — step-by-step solution | Explain MAths

area 2x^2,2x+6,-1,0

1) The area between

2 x^{2}

and

2 x + 6

on

[-1, 0]

is the integral of the absolute value of their difference

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

On

[-1, 0]

,

2 x + 6

is above

2 x^{2}

: integrate

- 2 x^{2} + 2 x + 6

⇥1.1) Let's integrate:

\int(- 2 x^{2} + 2 x + 6)dx

⇥

- \int(2 x^{2})dx + \int(2 x)dx + \int(6)dx

⇥ℹ

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

=

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

‖

\int(2 x)dx

=

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

‖

\int(6)dx

=

6 x

⇥

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

⇥ℹ

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

=

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

‖

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

=

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

‖

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

=

x^{2}

⇥

- \frac{2 x^{3}}{3} + x^{2} + 6 x

⇥1.2) Let's evaluate over the interval

[-1, 0]

:

(- \frac{2 \cdot 0^{3}}{3} + 0^{2} + 6 \cdot 0) - (- \frac{2 \cdot {(-1)}^{3}}{3} + {(-1)}^{2} + 6 \cdot (-1))

⇥ℹ

2 \cdot 0

=

0

‖

6 \cdot 0

=

0

‖

- (- \frac{2 \cdot {(-1)}^{3}}{3} + {(-1)}^{2} + 6 \cdot (-1))

=

\frac{2 \cdot {(-1)}^{3}}{3} - {(-1)}^{2} - 6 \cdot (-1)

⇥

0

+ 0

+ \frac{2 \cdot {(-1)}^{3}}{3}

- {(-1)}^{2}

- 6 \cdot (-1)

⇥ℹ

{(-1)}^{3}

=

-1

‖

2 \cdot (-1)

=

-2

‖

- {(-1)}^{2}

=

-1

‖

-6 \cdot (-1)

=

6

⇥

- \frac{2}{3} - 1 + 6

⇥

\frac{- 2 - 1 \cdot 3}{3} + 6

⇥

\frac{- 2 - 3}{3} + 6

⇥

- \frac{5}{3} + 6

⇥

\frac{- 5 + 6 \cdot 3}{3}

⇥

\frac{- 5 + 18}{3}

⇥

\frac{13}{3}

The area on

[-1, 0]

is

\frac{13}{3}

▶Area:

\frac{13}{3}