# Finding exact area intergration?

## Evaluate int_-1^4 x(3 − x) dx. b) Find the exact area of the shaded region in the figure.

Apr 23, 2018

$0.83 \text{ sq. units}$ (approx.)

#### Explanation:

We have,

${\int}_{-} {1}^{4} x \left(3 - x\right) \mathrm{dx}$

$= {\int}_{-} {1}^{4} \left(3 x - {x}^{2}\right) \mathrm{dx}$

$= {\int}_{-} {1}^{4} 3 x \mathrm{dx} - {\int}_{-} {1}^{4} {x}^{2} \mathrm{dx}$

Let's Evaluate ${\int}_{-} {1}^{4} 3 x \mathrm{dx}$ first.

So,
$= {\int}_{-} {1}^{4} 3 x \mathrm{dx} = {\left[\frac{3}{2} {x}^{2}\right]}_{-} {1}^{4} = \frac{3}{2} {4}^{2} - \frac{3}{2} {\left(- 1\right)}^{2} = 24 - \frac{3}{2} = \frac{48 - 3}{2} = \frac{45}{2}$

And, ${\int}_{-} {1}^{4} {x}^{2} \mathrm{dx} = {\left[\frac{1}{3} {x}^{3}\right]}_{-} {1}^{4} = \frac{1}{3} {4}^{3} - \frac{1}{3} {\left(- 1\right)}^{3} = \frac{64}{3} + \frac{1}{3} = \frac{65}{3}$

So, The Actual Integral :-

${\int}_{-} {1}^{4} x \left(3 - x\right) \mathrm{dx}$

$= \frac{45}{2} - \frac{65}{3} = \frac{135 - 130}{6} = \frac{5}{6} = 0.83$ (approx)

so, The Area Under the Curve = $0.83 \text{ sq.units}$.

Hope this helps.

Apr 24, 2018

The area is $= \frac{49}{6}$

#### Explanation:

The area is

${\int}_{1}^{4} \left(3 x - {x}^{2}\right) \mathrm{dx} = | {\int}_{1}^{0} \left(3 x - {x}^{2}\right) \mathrm{dx} | + {\int}_{0}^{3} \left(3 x - {x}^{2}\right) \mathrm{dx} + | {\int}_{3}^{4} \left(3 x - {x}^{2}\right) \mathrm{dx} |$

$= | {\left[\frac{3}{2} {x}^{2} - \frac{1}{3} {x}^{3}\right]}_{-} {1}^{0} | + {\left[\frac{3}{2} {x}^{2} - \frac{1}{3} {x}^{3}\right]}_{0}^{3} + | {\left[\frac{3}{2} {x}^{2} - \frac{1}{3} {x}^{3}\right]}_{3}^{4} |$

$= | \left(0\right) - \left(\frac{3}{2} + \frac{1}{3}\right) | + \left(\frac{27}{2} - 9\right) + | \left(24 - \frac{64}{3}\right) - \left(\frac{27}{2} - 9\right) |$

$= \frac{11}{6} + \frac{9}{2} + \frac{11}{6}$

$= \frac{49}{6}$

Apr 24, 2018

The total "actual area" is $\frac{49}{6}$

#### Explanation:

We seek that "actual" area bounded by the curve $x \left(3 - x\right)$ between $x = - 1$ and $x = 4$ as shown in the picture:

We will split this into $3$ intervals (actually we could split into $2$ and exploit the symmetry of the problem).

First, for convenience, consider the general case:

 I(a,b) = int_a^b \ x(3 − x) \ dx

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\int}_{a}^{b} \setminus 3 x - {x}^{2} \setminus \mathrm{dx}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\left[\frac{3}{2} {x}^{2} - \frac{1}{3} {x}^{3}\right]}_{a}^{b} \setminus$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{3}{2} \left({b}^{2} - {a}^{2}\right) - \frac{1}{3} \left({b}^{3} - {a}^{3}\right)$

Then we consider the three regions separately:

$I \left(- 1 , 0\right) = \frac{3}{2} \left(0 - 1\right) - \frac{1}{3} \left(0 + 1\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - \frac{3}{2} - \frac{1}{3} = - \frac{11}{6}$

$I \left(0 , 3\right) = \frac{3}{2} \left(9 - 0\right) - \frac{1}{3} \left(27 - 0\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{27}{2} - 9 = \frac{9}{2}$

$I \left(3 , 4\right) = \frac{3}{2} \left(16 - 9\right) - \frac{1}{3} \left(64 - 27\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{21}{2} - \frac{37}{3} = - \frac{11}{6}$ (expected via symmetry)

Then the total "actual area" is:

$A = | I \left(- 1 , 0\right) | + | I \left(0 , 3\right) | + | I \left(3 , 4\right) |$
$\setminus \setminus \setminus = | - \frac{11}{6} | + | \frac{9}{2} | + | - \frac{11}{6} |$
$\setminus \setminus \setminus = \frac{11}{6} + \frac{9}{2} + \frac{11}{6}$
$\setminus \setminus \setminus = \frac{49}{6}$

Note this is not the same as the "net" area, where area below the axis is counted negatively, which results in the incorrect answer:

 I(-1,4) = int_(-1)^4 \ x(3 − x) \ dx = 5/6