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Math  /  Calculus

QuestionFind the value of the constant CC for which the integral ∫0∞(xx2+1−C3x+1)dx\int_{0}^{\infty}\left(\frac{x}{x^{2}+1}-\frac{C}{3 x+1}\right) d x converges. Evaluate the integral for this value of CC. C=□C=\square
Value of convergent integral == □\square
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Studdy Solution

STEP 1

1. The integral is improper due to the upper limit being infinity.
2. For the integral to converge, the integrand must approach zero as x approaches infinity.
3. We can use techniques of partial fractions and limits to evaluate the integral.

STEP 2

1. Determine the condition for convergence.
2. Find the value of C that satisfies the convergence condition.
3. Evaluate the integral using the found value of C.

STEP 3

To determine the condition for convergence, we need to examine the behavior of the integrand as x approaches infinity:
lim⁡x→∞(xx2+1−C3x+1) \lim_{x \to \infty} \left(\frac{x}{x^2+1} - \frac{C}{3x+1}\right)
For the integral to converge, this limit must equal zero.

STEP 4

Let's evaluate the limit of each term separately:
lim⁡x→∞xx2+1=lim⁡x→∞1x+1x=0 \lim_{x \to \infty} \frac{x}{x^2+1} = \lim_{x \to \infty} \frac{1}{x+\frac{1}{x}} = 0
lim⁡x→∞C3x+1=0 \lim_{x \to \infty} \frac{C}{3x+1} = 0
For the overall limit to be zero, we don't need any specific condition on C. The integral will converge for any finite value of C.

STEP 5

Since the integral converges for any finite C, we need to find the specific value of C that makes the integral evaluable. Let's try to evaluate the integral:
∫0∞(xx2+1−C3x+1)dx \int_{0}^{\infty}\left(\frac{x}{x^{2}+1}-\frac{C}{3x+1}\right) dx
We can split this into two integrals:
∫0∞xx2+1dx−C∫0∞13x+1dx \int_{0}^{\infty}\frac{x}{x^{2}+1}dx - C\int_{0}^{\infty}\frac{1}{3x+1}dx

STEP 6

Let's evaluate each integral:
For ∫0∞xx2+1dx \int_{0}^{\infty}\frac{x}{x^{2}+1}dx : Let u = x^2 + 1, du = 2x dx 12∫1∞1udu=12[ln⁡∣u∣]1∞=12(∞−0)=∞ \frac{1}{2}\int_{1}^{\infty}\frac{1}{u}du = \frac{1}{2}[\ln|u|]_{1}^{\infty} = \frac{1}{2}(\infty - 0) = \infty
For ∫0∞13x+1dx \int_{0}^{\infty}\frac{1}{3x+1}dx : Let u = 3x + 1, du = 3 dx 13∫1∞1udu=13[ln⁡∣u∣]1∞=13(∞−0)=∞ \frac{1}{3}\int_{1}^{\infty}\frac{1}{u}du = \frac{1}{3}[\ln|u|]_{1}^{\infty} = \frac{1}{3}(\infty - 0) = \infty

STEP 7

For the overall integral to converge, we need these infinities to cancel out. This means we need:
12−C3=0 \frac{1}{2} - \frac{C}{3} = 0
Solving for C: C=32 C = \frac{3}{2}
This is the value of C that makes the integral convergent.

STEP 8

Now that we have C = 3/2, let's evaluate the integral:
∫0∞(xx2+1−3/23x+1)dx \int_{0}^{\infty}\left(\frac{x}{x^{2}+1}-\frac{3/2}{3x+1}\right) dx
We'll use substitution for each part:
For ∫0∞xx2+1dx \int_{0}^{\infty}\frac{x}{x^{2}+1}dx : Let u = x^2 + 1, du = 2x dx 12∫1∞1udu=12[ln⁡∣u∣]1∞=12[ln⁡(∞)−ln⁡(1)]=12ln⁡(∞) \frac{1}{2}\int_{1}^{\infty}\frac{1}{u}du = \frac{1}{2}[\ln|u|]_{1}^{\infty} = \frac{1}{2}[\ln(\infty) - \ln(1)] = \frac{1}{2}\ln(\infty)
For ∫0∞13x+1dx \int_{0}^{\infty}\frac{1}{3x+1}dx : Let u = 3x + 1, du = 3 dx 13∫1∞1udu=13[ln⁡∣u∣]1∞=13[ln⁡(∞)−ln⁡(1)]=13ln⁡(∞) \frac{1}{3}\int_{1}^{\infty}\frac{1}{u}du = \frac{1}{3}[\ln|u|]_{1}^{\infty} = \frac{1}{3}[\ln(\infty) - \ln(1)] = \frac{1}{3}\ln(\infty)

STEP 9

Combining the results:
12ln⁡(∞)−32⋅13ln⁡(∞)=12ln⁡(∞)−12ln⁡(∞)=0 \frac{1}{2}\ln(\infty) - \frac{3}{2} \cdot \frac{1}{3}\ln(\infty) = \frac{1}{2}\ln(\infty) - \frac{1}{2}\ln(\infty) = 0
Therefore, the value of the convergent integral is 0.
The value of C is: C=32 \boxed{C = \frac{3}{2}}
The value of the convergent integral is: 0 \boxed{0}

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