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Math  /  Calculus

Question21. Find the area of the region under the graph of the function f(x)=8x−x2f(x) = 8x - x^2 on the interval [0,2][0, 2].

Studdy Solution

STEP 1

What is this asking? We need to find the area underneath the curve f(x)=8x−x2f(x) = 8x - x^2 between x=0x = 0 and x=2x = 2. Watch out! Don't forget that the area *under* a curve means the area between the curve and the x-axis.
Also, make sure we're looking at the correct interval!

STEP 2

1. Define the function and interval
2. Calculate the definite integral
3. Simplify the result

STEP 3

Alright, so we've got our **function** f(x)=8x−x2f(x) = 8x - x^2.
We're looking at the **interval** from x=0x = 0 to x=2x = 2.
This means we're finding the definite integral of f(x)f(x) from **0** to **2**.

STEP 4

The definite integral represents the area under the curve.
Let's **calculate** it!
We'll use the power rule for integration, which says ∫xn dx=xn+1n+1+C\int x^n \, dx = \frac{x^{n+1}}{n+1} + C, where CC is the constant of integration.
Since this is a *definite* integral, the constant of integration will add to zero when we subtract the values of the integral at the bounds.

STEP 5

So, we have: ∫02(8x−x2) dx \int_{0}^{2} (8x - x^2) \, dx Let's integrate term by term: ∫028x dx−∫02x2 dx \int_{0}^{2} 8x \, dx - \int_{0}^{2} x^2 \, dx Applying the power rule: [8x1+11+1]02−[x2+12+1]02 \left[ \frac{8x^{1+1}}{1+1} \right]_0^2 - \left[ \frac{x^{2+1}}{2+1} \right]_0^2 [8x22]02−[x33]02 \left[ \frac{8x^2}{2} \right]_0^2 - \left[ \frac{x^3}{3} \right]_0^2 [4x2]02−[x33]02 \left[ 4x^2 \right]_0^2 - \left[ \frac{x^3}{3} \right]_0^2

STEP 6

Now, we **evaluate** at the **upper bound** (x=2x = 2) and **subtract** the evaluation at the **lower bound** (x=0x = 0): (4(2)2−4(0)2)−((2)33−(0)33) \left( 4(2)^2 - 4(0)^2 \right) - \left( \frac{(2)^3}{3} - \frac{(0)^3}{3} \right) (4⋅4−0)−(83−0) \left( 4 \cdot 4 - 0 \right) - \left( \frac{8}{3} - 0 \right) 16−83 16 - \frac{8}{3}

STEP 7

To subtract the fraction, let's get a common denominator.
We can rewrite 1616 as 161\frac{16}{1} and multiply by 33\frac{3}{3} to get a common denominator of 33: 161⋅33−83 \frac{16}{1} \cdot \frac{3}{3} - \frac{8}{3} 483−83 \frac{48}{3} - \frac{8}{3} 48−83 \frac{48-8}{3} 403 \frac{40}{3}

STEP 8

The area under the graph of f(x)=8x−x2f(x) = 8x - x^2 on the interval [0,2][0, 2] is 403\frac{40}{3}.

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