area y=6x,y=3x^2 — step-by-step solution

1) Consider the two curves with equations

y = 6 x

and

y = 3 x^{2}

At the intersections,

6 x

3 x^{2}

. Subtract the second expression from the first:

6 x - 3 x^{2} = 0

⇥1.1) Let's solve in

ℝ

6 x - 3 x^{2} = 0

⇥You can factor:

6 x - 3 x^{2}

into:

x \cdot (6 - 3 x)

⇥A product is zero if and only if one of its factors is zero: we solve each factor equal to 0

⇥⇥1.1.1) Let's solve in

ℝ

x = 0

⇥⇥Solution obtained:

x = 0

⇥⇥1.1.2) Let's solve in

ℝ

6 - 3 x = 0

⇥⇥

3 x = 6

⇥⇥ⓘ Both sides are divided by

3

⇥⇥

x = \frac{6}{3}

⇥⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

3

, their greatest common divisor:

\frac{6}{3}

2

⇥⇥Solution found:

x = 2

The area between

6 x

and

3 x^{2}

[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]

6 x

is above

3 x^{2}

: integrate

6 x - 3 x^{2}

⇥1.2) Let's compute the integral:

\int_{0}^{2} (6 x - 3 x^{2}) d x

⇥ⅰ

\int (6 x - 3 x^{2}) d x

\int 6 x d x - \int 3 x^{2} d x

⇥

\int 6 x d x - \int 3 x^{2} d x

⇥ⓘ A constant factor stays in front of the integral. An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

\int 6 x d x

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

- \int 3 x^{2} d x

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

⇥

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

⇥ⓘ Multiplying a fraction by a number multiplies its numerator:

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

\frac{6 x^{2}}{2}

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

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

■A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{6 x^{2}}{2}

3 x^{2}

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

- x^{3}

⇥

3 x^{2} - x^{3}

⇥This expression can be factorized into:

x^{2} \cdot (3 - x)

▷

x^{2} \cdot (3 - x)

⇥1.3) Let's evaluate over the interval

[0, 2]

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

⇥ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

2^{2}

2 \cdot 2

4

■Parentheses are computed first:

3 - 2

1

■Multiplying by 1 changes nothing:

4 \cdot 1

4

■Adding 0 changes nothing:

3 + 0

3

■A product with a zero factor is zero:

0 \cdot 3

0

⇥

4 + 0

⇥ⓘ Adding 0 changes nothing:

4 + 0

4

⇥

4

The area on

[0, 2]

4

▶Area:

4