integral from 0 to 4 of 4y-y^2 — step-by-step solution

1) Let's compute the integral:

\int_{0}^{4} (4 y - y^{2}) d y

ⅰ

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

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

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

ⓘ 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 y d y

4 \cdot \frac{y^{2}}{2}

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

- \int y^{2} d y

- \frac{y^{3}}{3}

4 \cdot \frac{y^{2}}{2} - \frac{y^{3}}{3}

ⓘ Multiplying a fraction by a number multiplies its numerator:

4 \cdot \frac{y^{2}}{2}

\frac{4 y^{2}}{2}

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

\frac{4 y^{2}}{2}

2 y^{2}

▷

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

2) Let's evaluate for the interval

[0, 4]

(2 \cdot 4^{2} - \frac{4^{3}}{3}) - (2 \cdot 0^{2} - \frac{0^{3}}{3})

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

4^{2}

4 \cdot 4

16

4^{3}

4 \cdot 4 \cdot 4

64

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

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

- 2 \cdot 0^{2} + \frac{0^{3}}{3}

2 \cdot 16

- \frac{64}{3}

- 2 \cdot 0^{2}

+ \frac{0^{3}}{3}

ⓘ Multiplications are done before additions and subtractions:

2 \cdot 16

2 \cdot (10 + 6)

2 \cdot 10 + 2 \cdot 6

20 + 12

32

■A product with a zero factor is zero:

-2 \cdot 0

0

32

- \frac{64}{3}

+ 0

+ 0

ⓘ Adding 0 changes nothing:

- \frac{64}{3} + 0 + 0

- \frac{64}{3}

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

32 - \frac{64}{3}

\frac{3 \cdot 32 - 64}{3}

\frac{3 \cdot 32 - 64}{3}

ⓘ Multiplications are done before additions and subtractions:

3 \cdot 32

3 \cdot (30 + 2)

3 \cdot 30 + 3 \cdot 2

90 + 6

96

\frac{96 - 64}{3}

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

96 - 64

32

▶Integral:

\frac{32}{3}

3) The numeric value of:

\frac{32}{3}

equals:

▷

\approx 10.6667