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Math  /  Calculus

QuestionEvaluate the definite integral ∫24x(x−2)3dx\int_{2}^{4} x(x-2)^{3} d x (Enter a numerical value. Round your answer to 2 decimal places as needed.)

Studdy Solution

STEP 1

What is this asking? We need to find the definite integral of x(x−2)3x(x-2)^3 from x=2x = 2 to x=4x = 4, which represents the **signed area** under the curve. Watch out! Don't forget to apply the **power rule** correctly after expanding the expression and be careful with the **bounds of integration** when evaluating the definite integral.

STEP 2

1. Expand the integrand
2. Find the indefinite integral
3. Evaluate the definite integral

STEP 3

Let's first expand (x−2)3(x-2)^3.
Remember, this is the same as (x−2)(x−2)(x−2)(x-2)(x-2)(x-2).
We can do this in stages.
First: (x−2)(x−2)=x(x−2)−2(x−2)=x2−2x−2x+4=x2−4x+4.(x-2)(x-2) = x(x-2) - 2(x-2) = x^2 - 2x - 2x + 4 = x^2 - 4x + 4. Then, multiply this result by (x−2)(x-2): (x2−4x+4)(x−2)=x(x2−4x+4)−2(x2−4x+4)=x3−4x2+4x−2x2+8x−8=x3−6x2+12x−8.(x^2 - 4x + 4)(x-2) = x(x^2 - 4x + 4) - 2(x^2 - 4x + 4) = x^3 - 4x^2 + 4x - 2x^2 + 8x - 8 = x^3 - 6x^2 + 12x - 8.

STEP 4

Now, we multiply the expanded expression by xx: x(x3−6x2+12x−8)=x4−6x3+12x2−8x.x(x^3 - 6x^2 + 12x - 8) = x^4 - 6x^3 + 12x^2 - 8x. So, our integral becomes: ∫24(x4−6x3+12x2−8x) dx.\int_{2}^{4} (x^4 - 6x^3 + 12x^2 - 8x) \, dx.

STEP 5

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 n≠−1n \neq -1).
Applying this to each term: ∫(x4−6x3+12x2−8x) dx=x55−6x44+12x33−8x22+C.\int (x^4 - 6x^3 + 12x^2 - 8x) \, dx = \frac{x^5}{5} - \frac{6x^4}{4} + \frac{12x^3}{3} - \frac{8x^2}{2} + C. Simplifying the fractions, we get: x55−3x42+4x3−4x2+C.\frac{x^5}{5} - \frac{3x^4}{2} + 4x^3 - 4x^2 + C.

STEP 6

Let's plug in our **upper bound**, x=4x = 4: 455−3⋅442+4⋅43−4⋅42=10245−7682+256−64=10245−384+192=10245−192.\frac{4^5}{5} - \frac{3 \cdot 4^4}{2} + 4 \cdot 4^3 - 4 \cdot 4^2 = \frac{1024}{5} - \frac{768}{2} + 256 - 64 = \frac{1024}{5} - 384 + 192 = \frac{1024}{5} - 192.

STEP 7

Now, let's plug in our **lower bound**, x=2x = 2: 255−3⋅242+4⋅23−4⋅22=325−482+32−16=325−24+16=325−8.\frac{2^5}{5} - \frac{3 \cdot 2^4}{2} + 4 \cdot 2^3 - 4 \cdot 2^2 = \frac{32}{5} - \frac{48}{2} + 32 - 16 = \frac{32}{5} - 24 + 16 = \frac{32}{5} - 8.

STEP 8

Finally, subtract the result from the lower bound from the result of the upper bound: (10245−192)−(325−8)=1024−325−192+8=9925−184=9925−184⋅55=992−9205=725=14.4. \left( \frac{1024}{5} - 192 \right) - \left( \frac{32}{5} - 8 \right) = \frac{1024 - 32}{5} - 192 + 8 = \frac{992}{5} - 184 = \frac{992}{5} - \frac{184 \cdot 5}{5} = \frac{992 - 920}{5} = \frac{72}{5} = 14.4.

STEP 9

The definite integral is **14.4**.

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