integral from -1 to 1 of 4-4x^2 — step-by-step solution

1) Let's integrate:

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

ⅰ

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

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

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

ⓘ The derivative of

4 x

4

, so

4 x

is an antiderivative of

4

\int 4 d x

4 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 4 x^{2} d x

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

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

ⓘ Multiplying a fraction by a number multiplies its numerator:

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

\frac{4 x^{3}}{3}

▷

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

2) Let's evaluate for the interval

[-1, 1]

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

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

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

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

4

- \frac{4}{3}

- 4 \cdot (-1)

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

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

4 - \frac{4}{3}

\frac{3 \cdot 4 - 4}{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

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

{(-1)}^{3}

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

-1

■Multiplying by -1 gives the opposite:

4 \cdot (-1)

-4

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

ⓘ Multiplications are done before additions and subtractions:

3 \cdot 4

12

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

4 - \frac{4}{3}

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

\frac{12 - 4}{3} + \frac{3 \cdot 4 - 4}{3}

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

12 - 4

8

■Multiplications are done before additions and subtractions:

3 \cdot 4

12

\frac{8}{3} + \frac{12 - 4}{3}

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

12 - 4

8

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

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

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

\frac{8 + 8}{3}

\frac{8 + 8}{3}

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

8 + 8

16

▶Integral:

\frac{16}{3}

3) The numeric value of:

\frac{16}{3}

is:

▷

\approx 5.3333