Math / Calculus

QuestionEvaluate. $\int_{1}^{e^{3}}\left(3 x+\frac{1}{x}\right) d x$

Studdy Solution

STEP 1

What is this asking? We need to find the definite integral of a function involving $x$ and $\frac{1}{x}$ from $1$ to $e^3$ . Watch out! Don't forget to apply the limits of integration correctly after finding the indefinite integral.
Also, remember the integral of $\frac{1}{x}$ is $\ln|x|$ , but since our bounds are positive, we can just use $\ln(x)$ .

STEP 2

1. Find the indefinite integral.
2. Evaluate the definite integral.

STEP 3

We're given the definite integral $\int_{1}^{e^{3}}\left(3x+\frac{1}{x}\right)dx$ .
Let's first find the indefinite integral, meaning we'll ignore the bounds for now.

STEP 4

The integral of $3x$ can be found using the power rule for integration: $\int ax^n dx = \frac{a}{n+1}x^{n+1} + C$ .
Here, $a = 3$ and $n = 1$ , so we have $\int 3x dx = \frac{3}{1+1}x^{1+1} + C = \frac{3}{2}x^2 + C$ .

STEP 5

The integral of $\frac{1}{x}$ is $\ln|x|$ .
Since our bounds of integration are positive, we can just write $\ln(x)$ .
So, $\int \frac{1}{x} dx = \ln(x) + C$ .

STEP 6

Putting it all together, the indefinite integral is $\int \left(3x + \frac{1}{x}\right) dx = \frac{3}{2}x^2 + \ln(x) + C$ .
Remember that $C$ is the constant of integration.

STEP 7

Now, let's evaluate the definite integral using the Fundamental Theorem of Calculus.
This tells us to plug in the upper limit, $e^3$ , into our indefinite integral and then subtract the result of plugging in the lower limit, $1$ .

STEP 8

Plugging in $e^3$ , we get $\frac{3}{2}(e^3)^2 + \ln(e^3) = \frac{3}{2}e^6 + 3$ .
Remember that $\ln(e^x) = x$ .

STEP 9

Plugging in $1$ , we get $\frac{3}{2}(1)^2 + \ln(1) = \frac{3}{2} + 0 = \frac{3}{2}$ .
Remember that $\ln(1) = 0$ .

STEP 10

Subtracting the result from the lower limit from the result of the upper limit gives us $\left(\frac{3}{2}e^6 + 3\right) - \frac{3}{2} = \frac{3}{2}e^6 + 3 - \frac{3}{2} = \frac{3}{2}e^6 + \frac{3}{2}$ .

STEP 11

Our final answer is $\frac{3}{2}e^6 + \frac{3}{2}$ .

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