integral from 0 to 2 of (4x^3-3x^2+2x)

Let's integrate:

\int_{0}^{2} (4 x^{3} - 3 x^{2} + 2 x) d x

ⅰ

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

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

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

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

- \int 3 x^{2} d x

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

\int 2 x d x

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

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

ⓘ Multiplying a fraction by a number multiplies its numerator:

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

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

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

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

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

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

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

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

x^{4}

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

- x^{3}

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

x^{2}

▷

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

Let's evaluate for the interval

[0, 2]

(2^{4} - 2^{3} + 2^{2}) - (0^{4} - 0^{3} + 0^{2})

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

2^{4}

2 \cdot 2 \cdot 2 \cdot 2

16

2^{3}

2 \cdot 2 \cdot 2

8

2^{2}

2 \cdot 2

4

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

- (0^{4} - 0^{3} + 0^{2})

- 0^{4} + 0^{3} - 0^{2}

16

- 8

+ 4

- 0^{4}

+ 0^{3}

- 0^{2}

ⓘ Start from

16

and move back

8

on the number line:

16 - 8

8

■Adding 0 changes nothing:

4

+ 0

+ 0

+ 0

4

8 + 4

ⓘ Start from

8

and move forward

4

on the number line:

8 + 4

12

▶Integral:

12