Math / Calculus

QuestionFind the area of the shaded region. $f(x)=20 x+x^{2}-x^{3}, g(x)=0$

Studdy Solution

STEP 1

What is this asking? Find the area trapped between the x-axis and a curvy line described by $f(x) = 20x + x^2 - x^3$ . Watch out! Make sure to find where the curve crosses the x-axis to get the boundaries for the area calculation.
Also, remember that area below the x-axis counts as negative, so we'll need to take the absolute value to get the total shaded area.

STEP 2

1. Find the x-intercepts
2. Set up the integral
3. Calculate the area

STEP 3

We start by finding where $f(x)$ crosses the x-axis.
This happens when $f(x) = 0$ .
So, we set $20x + x^2 - x^3 = 0$ .
Let's factor out an $x$ : $x(20 + x - x^2) = 0$ Now, we can factor the quadratic part: $-x(x^2 - x - 20) = 0$ $-x(x - 5)(x + 4) = 0$

STEP 4

This gives us three roots: $x = 0$ , $x = 5$ , and $x = -4$ .
These are the points where our curve intersects the x-axis!

STEP 5

Since the curve dips below the x-axis, we'll have to split the integral into two parts: one from $x = -4$ to $x = 0$ and another from $x = 0$ to $x = 5$ .
This is because the area under the curve is negative when the curve is below the x-axis.

STEP 6

The first integral represents the area from $x = -4$ to $x = 0$ : $\int_{-4}^{0} (20x + x^2 - x^3) \, dx$ The second integral represents the area from $x = 0$ to $x = 5$ : $\int_{0}^{5} (20x + x^2 - x^3) \, dx$

STEP 7

Let's evaluate the first integral: $\int_{-4}^{0} (20x + x^2 - x^3) \, dx = \left[10x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4\right]_{-4}^{0}$ Plugging in the limits of integration, we get: $(0) - \left(10(-4)^2 + \frac{1}{3}(-4)^3 - \frac{1}{4}(-4)^4\right) = - \left(160 - \frac{64}{3} - 64\right) = -\frac{224}{3}$ Since the area is negative, we take the absolute value: $\frac{224}{3}$ .

STEP 8

Now, let's evaluate the second integral: $\int_{0}^{5} (20x + x^2 - x^3) \, dx = \left[10x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4\right]_{0}^{5}$ Plugging in the limits of integration, we get: $\left(10(5)^2 + \frac{1}{3}(5)^3 - \frac{1}{4}(5)^4\right) - (0) = 250 + \frac{125}{3} - \frac{625}{4} = \frac{625}{12}$

STEP 9

Finally, we add the two areas together: $\frac{224}{3} + \frac{625}{12} = \frac{912}{12} + \frac{625}{12} = \frac{1517}{12}$

STEP 10

The total area of the shaded region is $\frac{1517}{12}$ .

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