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Math  /  Calculus

QuestionThe answer above is NOT correct. Find the area between the curves: y=x3−13x2+40xy=x^{3}-13 x^{2}+40 x and y=−x3+13x2−40xy=-x^{3}+13 x^{2}-40 x □\square

Studdy Solution

STEP 1

What is this asking? We need to find the area trapped between two curvy functions, one of which is just the negative of the other! Watch out! Make sure to find where the curves intersect to get the bounds of integration.
Also, don't forget that area is *always* positive!

STEP 2

1. Find Intersection Points
2. Set Up the Integral
3. Evaluate the Integral

STEP 3

Let's **find** where these two crazy curves meet!
We set the two equations equal to each other: x3−13x2+40x=−x3+13x2−40x x^3 - 13x^2 + 40x = -x^3 + 13x^2 - 40x

STEP 4

Time to **simplify**!
Add x3−13x2+40xx^3 - 13x^2 + 40x to both sides: 2x3−26x2+80x=0 2x^3 - 26x^2 + 80x = 0

STEP 5

**Factor** out 2x2x: 2x(x2−13x+40)=0 2x(x^2 - 13x + 40) = 0

STEP 6

Now, let's **factor** the quadratic.
We're looking for two numbers that multiply to **40** and add to **-13**.
How about **-5** and **-8**? 2x(x−5)(x−8)=0 2x(x-5)(x-8) = 0

STEP 7

So, our intersection points are x=0x = 0, x=5x = 5, and x=8x = 8!

STEP 8

Since y=x3−13x2+40xy = x^3 - 13x^2 + 40x is greater than y=−x3+13x2−40xy = -x^3 + 13x^2 - 40x between x=0x=0 and x=5x=5 (and the other way around between x=5x=5 and x=8x=8), we'll integrate the *absolute difference* between the two functions.
This ensures we're always calculating a positive area!

STEP 9

Our **integral** will look like this: ∫08∣(x3−13x2+40x)−(−x3+13x2−40x)∣ dx \int_{0}^{8} |(x^3 - 13x^2 + 40x) - (-x^3 + 13x^2 - 40x)| \, dx

STEP 10

**Simplify** the integrand: ∫08∣2x3−26x2+80x∣ dx \int_{0}^{8} |2x^3 - 26x^2 + 80x| \, dx

STEP 11

Since the expression inside the absolute value changes sign at x=5x=5, we split the integral into two parts: ∫05(2x3−26x2+80x) dx−∫58(2x3−26x2+80x) dx \int_{0}^{5} (2x^3 - 26x^2 + 80x) \, dx - \int_{5}^{8} (2x^3 - 26x^2 + 80x) \, dx

STEP 12

**Find the antiderivative**: [12x4−263x3+40x2]05−[12x4−263x3+40x2]58 \left[ \frac{1}{2}x^4 - \frac{26}{3}x^3 + 40x^2 \right]_0^5 - \left[ \frac{1}{2}x^4 - \frac{26}{3}x^3 + 40x^2 \right]_5^8

STEP 13

**Evaluate** at the bounds: (6252−32503+1000)−(0)−[(2048−133123+2560)−(6252−32503+1000)] \left( \frac{625}{2} - \frac{3250}{3} + 1000 \right) - (0) - \left[ \left( 2048 - \frac{13312}{3} + 2560 \right) - \left( \frac{625}{2} - \frac{3250}{3} + 1000 \right) \right]

STEP 14

**Simplify**: 18756−(76806−18756)=18756−58056=−39306 \frac{1875}{6} - \left( \frac{7680}{6} - \frac{1875}{6} \right) = \frac{1875}{6} - \frac{5805}{6} = -\frac{3930}{6} Since area must be positive, we take the absolute value: 39306=655\frac{3930}{6} = 655

STEP 15

The area between the curves is **655** square units.

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