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

1) Consider the two curves with equations

y = x^{2} - 2 x

and

y = x + 4

At the intersections,

x^{2} - 2 x

x + 4

. Subtract the second expression from the first:

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

= 0, giving

x^{2} - 3 x - 4 = 0

⇥1.1) Let's solve in

ℝ

x^{2} - 3 x - 4 = 0

⇥Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -3

and

c = -4

⇥

Δ = {(-3)}^{2} - 4 \cdot 1 \cdot (-4)

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

{(-3)}^{2}

(-3) \cdot (-3)

9

⇥

Δ = 9 - 4 \cdot (-4)

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

- 4 \cdot (-4)

16

⇥

Δ = 9 + 16

⇥ⓘ Start from

9

and move forward

16

on the number line:

9 + 16

25

⇥

Δ = 25

⇥Since

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-3) - \sqrt{25}}{2 \cdot 1}

⇥

x_{1} = \frac{3 - \sqrt{25}}{2}

⇥ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{25}

5

;

5^{2}

25

⇥

x_{1} = \frac{3 - 5}{2}

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

3 - 5

-2

⇥

x_{1} = - \frac{2}{2}

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

2

, their greatest common divisor:

- \frac{2}{2}

-1

⇥

x_{1} = -1

⇥The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-3) + \sqrt{25}}{2 \cdot 1}

⇥

x_{2} = \frac{3 + \sqrt{25}}{2}

⇥ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{25}

5

;

5^{2}

25

⇥

x_{2} = \frac{3 + 5}{2}

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

3 + 5

8

⇥

x_{2} = \frac{8}{2}

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

2

, their greatest common divisor:

\frac{8}{2}

4

⇥

x_{2} = 4

⇥Solutions obtained:

x = {-1, 4}

The area between

x^{2} - 2 x

and

x + 4

[-1, 4]

is the integral of the absolute value of their difference

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

[-1, 4]

x + 4

is above

x^{2} - 2 x

: integrate

- x^{2} + 3 x + 4

⇥1.2) Let's compute the integral:

\int_{-1}^{4} (- x^{2} + 3 x + 4) d x

⇥ⅰ

\int (- x^{2} + 3 x + 4) d x

- \int x^{2} d x + \int 3 x d x + \int 4 d x

⇥

- \int x^{2} d x + \int 3 x d x + \int 4 d x

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

- \int x^{2} d x

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

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

\int 3 x d x

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

■The derivative of

4 x

4

, so

4 x

is an antiderivative of

4

\int 4 d x

4 x

⇥

- \frac{x^{3}}{3} + 3 \cdot \frac{x^{2}}{2} + 4 x

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

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

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

▷

- \frac{x^{3}}{3} + \frac{3 x^{2}}{2} + 4 x

⇥1.3) Let's evaluate for the interval

[-1, 4]

(- \frac{4^{3}}{3} + \frac{3 \cdot 4^{2}}{2} + 4 \cdot 4) - (- \frac{{(-1)}^{3}}{3} + \frac{3 \cdot {(-1)}^{2}}{2} + 4 \cdot (-1))

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

4^{3}

4 \cdot 4 \cdot 4

64

4^{2}

4 \cdot 4

16

■Multiplications are done before additions and subtractions:

4 \cdot 4

16

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

- (- \frac{{(-1)}^{3}}{3} + \frac{3 \cdot {(-1)}^{2}}{2} + 4 \cdot (-1))

\frac{{(-1)}^{3}}{3} - \frac{3 \cdot {(-1)}^{2}}{2} - 4 \cdot (-1)

⇥

- \frac{64}{3}

+ \frac{3 \cdot 16}{2}

+ 16

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

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

- 4 \cdot (-1)

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

{(-1)}^{3}

(-1) \cdot (-1) \cdot (-1)

-1

{(-1)}^{2}

(-1) \cdot (-1)

1

■Multiplying by 1 changes nothing:

3 \cdot 1

3

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

-4 \cdot (-1)

4

⇥ⅰ

3 \cdot 16

3 \cdot (10 + 6)

3 \cdot 10 + 3 \cdot 6

30 + 18

48

⇥

- \frac{64}{3}

+ \frac{48}{2}

+ 16

- \frac{1}{3}

- \frac{3}{2}

+ 4

⇥ⓘ To add fractions, put them over the same denominator, then add the numerators. We choose

6

, the least common multiple of

3

and

2

. Multiplying the numerator and the denominator by the same number does not change the fraction:

- \frac{64}{3} + \frac{48}{2}

\frac{-64 \cdot 2}{3 \cdot 2} + \frac{48 \cdot 3}{2 \cdot 3}

\frac{- 64 \cdot 2 + 48 \cdot 3}{3 \cdot 2}

⇥

\frac{- 64 \cdot 2 + 48 \cdot 3}{3 \cdot 2}

+ 16

- \frac{1}{3}

- \frac{3}{2}

+ 4

⇥ⓘ Multiplications are done before additions and subtractions:

64 \cdot 2

(60 + 4) \cdot 2

60 \cdot 2 + 4 \cdot 2

120 + 8

128

48 \cdot 3

(40 + 8) \cdot 3

40 \cdot 3 + 8 \cdot 3

120 + 24

144

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

16 - \frac{1}{3}

\frac{3 \cdot 16 - 1}{3}

⇥ⅰ

3 \cdot 2

6

⇥

\frac{- 128 + 144}{6}

+ \frac{3 \cdot 16 - 1}{3}

- \frac{3}{2}

+ 4

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

- 128 + 144

16

■Multiplications are done before additions and subtractions:

3 \cdot 16

3 \cdot (10 + 6)

3 \cdot 10 + 3 \cdot 6

30 + 18

48

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

- \frac{3}{2} + 4

\frac{- 3 + 4 \cdot 2}{2}

⇥

\frac{16}{6} + \frac{48 - 1}{3} + \frac{- 3 + 4 \cdot 2}{2}

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

2

, their greatest common divisor:

\frac{16}{6}

\frac{8 \cdot 2}{3 \cdot 2}

\frac{8}{3}

■The numerator and the denominator are computed before the division:

48 - 1

47

■Multiplications are done before additions and subtractions:

4 \cdot 2

8

⇥

\frac{8}{3} + \frac{47}{3} + \frac{- 3 + 8}{2}

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

\frac{8}{3} + \frac{47}{3}

\frac{8 + 47}{3}

■The numerator and the denominator are computed before the division:

- 3 + 8

5

⇥

\frac{8 + 47}{3} + \frac{5}{2}

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

8 + 47

55

⇥

\frac{55}{3} + \frac{5}{2}

⇥ⓘ To add fractions, put them over the same denominator, then add the numerators. We choose

6

, the least common multiple of

3

and

2

. Multiplying the numerator and the denominator by the same number does not change the fraction:

\frac{55}{3} + \frac{5}{2}

\frac{55 \cdot 2}{3 \cdot 2} + \frac{5 \cdot 3}{2 \cdot 3}

\frac{55 \cdot 2 + 5 \cdot 3}{3 \cdot 2}

⇥

\frac{55 \cdot 2 + 5 \cdot 3}{3 \cdot 2}

⇥ⓘ Multiplications are done before additions and subtractions:

55 \cdot 2

(50 + 5) \cdot 2

50 \cdot 2 + 5 \cdot 2

100 + 10

110

5 \cdot 3

15

⇥ⅰ

3 \cdot 2

6

⇥

\frac{110 + 15}{6}

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

110 + 15

125

⇥

\frac{125}{6}

The area on

[-1, 4]

\frac{125}{6}

▶Area:

\frac{125}{6}

2) The numeric value of:

\frac{125}{6}

is:

▷

\approx 20.8333