area y=x+4,y=x^2+x-12 — step-by-step solution

1) Consider the two curves with equations

y = x + 4

and

y = x^{2} + x - 12

At the intersections,

x + 4

x^{2} + x - 12

. Subtract the second expression from the first:

(x + 4) - (x^{2} + x - 12)

= 0, giving

16 - x^{2} = 0

⇥1.1) Let's solve in

ℝ

16 - x^{2} = 0

⇥The two numbers whose square is

16

are

-4

and

4

⇥Solutions obtained:

x = {-4, 4}

The area between

x + 4

and

x^{2} + x - 12

[-4, 4]

is the integral of the absolute value of their difference

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

[-4, 4]

x + 4

is above

x^{2} + x - 12

: integrate

16 - x^{2}

⇥1.2) Let's compute the integral:

\int_{-4}^{4} (16 - x^{2}) d x

⇥ⅰ

\int (16 - x^{2}) d x

\int 16 d x - \int x^{2} d x

⇥

\int 16 d x - \int x^{2} d x

⇥ⓘ The derivative of

16 x

16

, so

16 x

is an antiderivative of

16

\int 16 d x

16 x

■An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

- \int x^{2} d x

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

▷

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

⇥1.3) Let's evaluate for the interval

[-4, 4]

(16 \cdot 4 - \frac{4^{3}}{3}) - (16 \cdot (-4) - \frac{{(-4)}^{3}}{3})

⇥ⓘ Multiplications are done before additions and subtractions:

16 \cdot 4

(10 + 6) \cdot 4

10 \cdot 4 + 6 \cdot 4

40 + 24

64

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

4^{3}

4 \cdot 4 \cdot 4

64

■A minus sign before a parenthesis changes the sign of each of its terms:

- (16 \cdot (-4) - \frac{{(-4)}^{3}}{3})

- 16 \cdot (-4) + \frac{{(-4)}^{3}}{3}

⇥

64

- \frac{64}{3}

- 16 \cdot (-4)

+ \frac{{(-4)}^{3}}{3}

⇥ⓘ To add fractions, put them over the same denominator, then add the numerators:

64 - \frac{64}{3}

\frac{3 \cdot 64 - 64}{3}

■Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

-16 \cdot (-4)

16 \cdot 4

(10 + 6) \cdot 4

10 \cdot 4 + 6 \cdot 4

40 + 24

64

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

{(-4)}^{3}

(-4) \cdot (-4) \cdot (-4)

-64

⇥

\frac{3 \cdot 64 - 64}{3} + 64 - \frac{64}{3}

⇥ⓘ Multiplications are done before additions and subtractions:

3 \cdot 64

3 \cdot (60 + 4)

3 \cdot 60 + 3 \cdot 4

180 + 12

192

■To add fractions, put them over the same denominator, then add the numerators:

64 - \frac{64}{3}

\frac{3 \cdot 64 - 64}{3}

⇥

\frac{192 - 64}{3} + \frac{3 \cdot 64 - 64}{3}

⇥ⓘ The numerator and the denominator are computed before the division:

192 - 64

128

■Multiplications are done before additions and subtractions:

3 \cdot 64

3 \cdot (60 + 4)

3 \cdot 60 + 3 \cdot 4

180 + 12

192

⇥

\frac{128}{3} + \frac{192 - 64}{3}

⇥ⓘ The numerator and the denominator are computed before the division:

192 - 64

128

⇥

\frac{128}{3} + \frac{128}{3}

⇥ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{128}{3} + \frac{128}{3}

\frac{128 + 128}{3}

⇥

\frac{128 + 128}{3}

⇥ⓘ The numerator and the denominator are computed before the division:

128 + 128

256

⇥

\frac{256}{3}

The area on

[-4, 4]

\frac{256}{3}

▶Area:

\frac{256}{3}

2) The numeric value of:

\frac{256}{3}

equals:

▷

\approx 85.3333