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Math  /  Calculus

QuestionEvaluate the definite integral. ∫16(x2−7x)dx\int_{1}^{6}\left(\frac{x}{2}-\frac{7}{x}\right) d x □\square

Studdy Solution

STEP 1

What is this asking? We need to find the definite integral of a function involving xx and 1x\frac{1}{x} from x=1x = 1 to x=6x = 6. Watch out! Don't forget to apply the upper and lower limits of integration correctly!
Also, remember the integral of 1x\frac{1}{x} is ln⁡∣x∣\ln|x|, not just ln⁡(x)\ln(x), but since our bounds are positive, we can just use ln⁡(x)\ln(x).

STEP 2

1. Rewrite the integral for clarity
2. Find the indefinite integral
3. Evaluate the definite integral

STEP 3

Let's **rewrite** the integral to make it easier to work with!
We can split the integral of a sum (or difference) into the sum (or difference) of integrals.
Also, we can pull out constant factors.
So, we can rewrite our integral like this: ∫16(x2−7x)dx=∫1612⋅x dx−∫167⋅1x dx=12∫16x dx−7∫161x dx \int_{1}^{6} \left( \frac{x}{2} - \frac{7}{x} \right) dx = \int_{1}^{6} \frac{1}{2} \cdot x \, dx - \int_{1}^{6} 7 \cdot \frac{1}{x} \, dx = \frac{1}{2} \int_{1}^{6} x \, dx - 7 \int_{1}^{6} \frac{1}{x} \, dx

STEP 4

Now, let's **find** the *indefinite integral*!
Remember, the power rule for integration says ∫xn dx=xn+1n+1+C\int x^n \, dx = \frac{x^{n+1}}{n+1} + C for n≠−1n \ne -1.
Also, ∫1xdx=ln⁡∣x∣+C\int \frac{1}{x} dx = \ln|x| + C.
In our case, since the bounds of integration are positive, we can just use ln⁡(x)\ln(x). ∫(x2−7x)dx=12∫x dx−7∫1x dx=12⋅x22−7ln⁡(x)+C=x24−7ln⁡(x)+C \int \left( \frac{x}{2} - \frac{7}{x} \right) dx = \frac{1}{2} \int x \, dx - 7 \int \frac{1}{x} \, dx = \frac{1}{2} \cdot \frac{x^2}{2} - 7 \ln(x) + C = \frac{x^2}{4} - 7\ln(x) + C

STEP 5

Let's **evaluate** the *definite integral* using the Fundamental Theorem of Calculus!
We'll plug in our **upper limit** of integration (x=6x = 6) and **lower limit** of integration (x=1x = 1) into our indefinite integral and subtract the results. [x24−7ln⁡(x)]16=(624−7ln⁡(6))−(124−7ln⁡(1)) \left[ \frac{x^2}{4} - 7\ln(x) \right]_1^6 = \left( \frac{6^2}{4} - 7\ln(6) \right) - \left( \frac{1^2}{4} - 7\ln(1) \right)

STEP 6

Remember that ln⁡(1)=0\ln(1) = 0, so we have: (364−7ln⁡(6))−(14−0)=9−7ln⁡(6)−14=364−14−7ln⁡(6)=354−7ln⁡(6) \left( \frac{36}{4} - 7\ln(6) \right) - \left( \frac{1}{4} - 0 \right) = 9 - 7\ln(6) - \frac{1}{4} = \frac{36}{4} - \frac{1}{4} - 7\ln(6) = \frac{35}{4} - 7\ln(6)

STEP 7

Our **final answer** is 354−7ln⁡(6)\frac{35}{4} - 7\ln(6).

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