Evaluate the definite integral.∫2e3+xxdx\int_2^{e}\frac{3+x}{x}dx

1 + e − 3 ln ⁡ 2 ≈ 1.64 1+e-3\ln2\thickapprox1.64

Evaluate the definite integral. ∫ 2 e 3 + x x d x \int_2^{e}\frac{3+x}{x}dx

Here we're asked to evaluate the definite integral from two to e of three plus x over x dx. Now looking at this integral, since I have two terms in my numerator and only one term in my denominator, that's usually a good clue that I should actually break this up into two separate fractions that will likely be easier to deal with than this whole fraction altogether. So I'm going to rewrite this as the integral from two to e of three over x plus x over x, just splitting that up into two fractions. Now we can see here that x over x, that's gonna cancel each other out and we're just left there with one. So this is the integral from two to e of three over x plus one d x. So now we can just go ahead and find our antiderivatives of each of these terms to evaluate this integral. So for focusing on that first term here, three over x, three is just a constant. So this is gonna be three times the natural log of the absolute value of x based on the rule that we learned earlier. Then the antiderivative of one is just x, so that's plus x and then bounded from two to e. So now that we have that, we just need to apply those bounds. So plugging my upper bound first here, that's three times the natural log of e. I don't have to worry about my absolute value bars here because e is already a positive value. Then plus e. So that's my upper bound plugged in. Then subtracting off my lower bound plugged in, this is three times the natural log of two. Again, two is already positive, so I don't have to worry about my absolute value plus two. Then we just wanna do some simplification here. The natural log of e is just one. So this is three plus e, and then minus three times the natural log of two. And then having distributed that negative, we are subtracting off two. I can go ahead and combine some like terms here, three and negative two is just gonna be one. So this is one plus e minus three natural log of two. Now if you aren't using a calculator in your course, then this is your fully simplified answer here and we are done. One plus e minus three natural log of two. But if you are expected to give a decimal answer and you type this into your calculator, you should ultimately get here 1.64 having rounded off to two decimals. Now if you have any questions, feel free to let us know in the comments, and I'll see you in the next one.

Evaluate the definite integral. ∫ 2 e 3 + x x d x \int_2^{e}\frac{3+x}{x}dx

1 + e − 3 ln ⁡ 2 ≈ 1.64 1+e-3\ln2\thickapprox1.64

4 e − 8 ≈ 2.87 4e-8\thickapprox2.87

e − 3 ln ⁡ 2 − 1 ≈ − 0.36 e-3\ln2-1\thickapprox-0.36

e − 3 ≈ − 0.28 e-3\thickapprox-0.28

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0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
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4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
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- Area Between Curves
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10. Physics Applications of Integrals 3h 16m
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11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
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16. Parametric Equations & Polar Coordinates7h 58m

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

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Multiple Choice

Evaluate the definite integral.
$\int_{2}^{e} \frac{3 + x}{x} d x$

$4 e - 8 \approx 2.87$

$e - 3 \ln 2 - 1 \approx - 0.36$

$e - 3 \approx - 0.28$

$1 + e - 3 \ln 2 \approx 1.64$

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Verified step by step guidance

Step 1: Recognize that the integral to evaluate is $ \int_2^e \frac{3+x}{x} \, dx $. This is a definite integral, and the integrand can be simplified by splitting the fraction $ \frac{3+x}{x} $ into two terms: $ \frac{3}{x} + 1 $.

Step 2: Rewrite the integral as $ \int_2^e \left( \frac{3}{x} + 1 \right) \, dx $. Now, split the integral into two separate integrals: $ \int_2^e \frac{3}{x} \, dx + \int_2^e 1 \, dx $.

Step 3: Evaluate the first integral, $ \int_2^e \frac{3}{x} \, dx $. The antiderivative of $ \frac{3}{x} $ is $ 3 \ln|x| $. Apply the limits of integration: $ 3 \ln|e| - 3 \ln|2| $. Since $ \ln|e| = 1 $, this simplifies to $ 3(1) - 3 \ln(2) $.

Step 4: Evaluate the second integral, $ \int_2^e 1 \, dx $. The antiderivative of $ 1 $ is $ x $. Apply the limits of integration: $ e - 2 $.

Step 5: Combine the results from the two integrals. The final expression is $ \left[ 3 - 3 \ln(2) \right] + \left[ e - 2 \right] $. Simplify this expression to get the final result.