Math  /  Calculus

Question(1 point) 14x5+3xdx=\int_{1}^{4} \frac{x^{5}+3}{x} d x= \square Preview My Answers Submit Answers

Studdy Solution

STEP 1

What is this asking? Calculate the definite integral of (x5+3)/x(x^5 + 3)/x from x=1x = 1 to x=4x = 4. Watch out! Don't forget to apply the limits of integration after finding the indefinite integral!

STEP 2

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

STEP 3

We can **rewrite** our integrand x5+3x\frac{x^5 + 3}{x} as the sum of two simpler fractions: x5x+3x\frac{x^5}{x} + \frac{3}{x}.
This makes it easier to integrate!

STEP 4

Now, let's **simplify** each term. x5x\frac{x^5}{x} simplifies to x4x^4 (since x5x^5 divided by xx is x51=x4x^{5-1} = x^4).
The second term, 3x\frac{3}{x}, stays as it is.
So, our simplified integrand is x4+3xx^4 + \frac{3}{x}.
Much nicer, right?

STEP 5

Let's **integrate** x4+3xx^4 + \frac{3}{x} with respect to xx.
Remember the power rule: the integral of xnx^n is xn+1n+1\frac{x^{n+1}}{n+1}.
Applying this to x4x^4, we get x4+14+1=x55\frac{x^{4+1}}{4+1} = \frac{x^5}{5}.
The integral of 1x\frac{1}{x} is lnx\ln|x|, so the integral of 3x\frac{3}{x} is 3lnx=3lnx3 \cdot \ln|x| = 3\ln|x|.
Don't forget to add the constant of integration, CC!
So, our indefinite integral is x55+3lnx+C\frac{x^5}{5} + 3\ln|x| + C.

STEP 6

Now, let's **evaluate** the definite integral from x=1x = 1 to x=4x = 4.
We'll substitute these values into our indefinite integral and subtract the result at the lower limit from the result at the upper limit.

STEP 7

Substituting x=4x = 4, we get 455+3ln4=10245+3ln4\frac{4^5}{5} + 3\ln|4| = \frac{1024}{5} + 3\ln 4.

STEP 8

Substituting x=1x = 1, we get 155+3ln1=15+30=15\frac{1^5}{5} + 3\ln|1| = \frac{1}{5} + 3 \cdot 0 = \frac{1}{5} (since ln1=0\ln 1 = 0).

STEP 9

Finally, we **subtract** the result at the lower limit from the result at the upper limit: (10245+3ln4)15=10235+3ln4(\frac{1024}{5} + 3\ln 4) - \frac{1}{5} = \frac{1023}{5} + 3\ln 4.
Since 4=224 = 2^2, we can rewrite 3ln43\ln 4 as 3ln(22)=32ln2=6ln23\ln(2^2) = 3 \cdot 2\ln 2 = 6\ln 2.
So, our **final result** is 10235+6ln2\frac{1023}{5} + 6\ln 2.

STEP 10

The definite integral is 10235+6ln2\frac{1023}{5} + 6\ln 2.

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