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Math  /  Geometry

QuestionFind the area of the shaded region. f(x)=20x+x2−x3,g(x)=0f(x)=20 x+x^{2}-x^{3}, g(x)=0

Studdy Solution

STEP 1

What is this asking? We need to find the area trapped between a curvy function and the x-axis! Watch out! We need to find where the function crosses the x-axis to get our bounds of integration.
Don't forget that area below the x-axis counts as negative!

STEP 2

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

STEP 3

Alright, let's **find** those **x-intercepts**!
We're looking for where  f(x)=20x+x2−x3=0\ f(x) = 20x + x^2 - x^3 = 0 .
Factoring out an xx gives us x(20+x−x2)=0 x(20 + x - x^2) = 0 .

STEP 4

One **x-intercept** is clearly x=0x = 0.
Now, let's deal with that quadratic 20+x−x2=0 20 + x - x^2 = 0 .
Multiplying by −1-1 to make it easier to factor, we get x2−x−20=0 x^2 - x - 20 = 0 .

STEP 5

This factors nicely into (x−5)(x+4)=0 (x-5)(x+4) = 0 , giving us x=5 x = 5 and x=−4 x = -4 as our other **intercepts**!
So, our **x-intercepts** are x=−4 x = -4 , x=0 x = 0 , and x=5 x = 5 .

STEP 6

Now, we need to **set up** our **integral** to find the **area**.
We'll have two integrals, one from x=−4 x = -4 to x=0 x = 0 and another from x=0 x = 0 to x=5 x = 5 .

STEP 7

Our **first integral** will be ∫−40(20x+x2−x3) dx \int_{-4}^{0} (20x + x^2 - x^3) \, dx .
Our **second integral** will be ∫05(20x+x2−x3) dx \int_{0}^{5} (20x + x^2 - x^3) \, dx .

STEP 8

Let's **evaluate** the **first integral**: ∫−40(20x+x2−x3) dx=[20x22+x33−x44]−40=[10x2+x33−x44]−40 \int_{-4}^{0} (20x + x^2 - x^3) \, dx = \left[ \frac{20x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} \right]_{-4}^{0} = \left[ 10x^2 + \frac{x^3}{3} - \frac{x^4}{4} \right]_{-4}^{0}

STEP 9

Plugging in our **bounds**, we get: (0)−(10(−4)2+(−4)33−(−4)44)=−(160−643−64)=−(96−643)=−288−643=−2243 \left( 0 \right) - \left( 10(-4)^2 + \frac{(-4)^3}{3} - \frac{(-4)^4}{4} \right) = - \left( 160 - \frac{64}{3} - 64 \right) = - \left( 96 - \frac{64}{3} \right) = - \frac{288 - 64}{3} = -\frac{224}{3}

STEP 10

Now, let's **tackle** the **second integral**: ∫05(20x+x2−x3) dx=[10x2+x33−x44]05 \int_{0}^{5} (20x + x^2 - x^3) \, dx = \left[ 10x^2 + \frac{x^3}{3} - \frac{x^4}{4} \right]_{0}^{5}

STEP 11

Plugging in our **bounds**, we have: (10(5)2+(5)33−(5)44)−(0)=250+1253−6254=3000+500−187512=162512 \left( 10(5)^2 + \frac{(5)^3}{3} - \frac{(5)^4}{4} \right) - \left( 0 \right) = 250 + \frac{125}{3} - \frac{625}{4} = \frac{3000 + 500 - 1875}{12} = \frac{1625}{12}

STEP 12

Since the first area is below the x-axis, we take its absolute value and add it to the second area: 2243+162512=896+162512=252112 \frac{224}{3} + \frac{1625}{12} = \frac{896 + 1625}{12} = \frac{2521}{12} .

STEP 13

The **total area** of the shaded region is 252112 \frac{2521}{12} .

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