Math

QuestionDoes the integral 031xdx\int_{0}^{3} \frac{1}{x} dx converge or diverge?

Studdy Solution

STEP 1

Assumptions1. We are asked to evaluate the improper integral 031xdx\int_{0}^{3} \frac{1}{x} d x. An integral is said to converge if the limit exists and is finite, and diverge if the limit does not exist or is infinite.

STEP 2

The given integral is an improper integral because it has a singularity at x=0. To evaluate this, we need to take the limit as the lower limit of integration approaches the singularity.
01xdx=lima0+a1xdx\int_{0}^{} \frac{1}{x} d x = \lim_{a \to0^{+}} \int_{a}^{} \frac{1}{x} d x

STEP 3

Now, we need to evaluate the integral. The antiderivative of 1x\frac{1}{x} is lnx\ln|x|.
lima0+a31xdx=lima0+[lnx]a3\lim_{a \to0^{+}} \int_{a}^{3} \frac{1}{x} d x = \lim_{a \to0^{+}} [\ln|x|]_{a}^{3}

STEP 4

Next, we substitute the limits of integration into the antiderivative.
lima0+[lnx]a3=lima0+(ln3lna)\lim_{a \to0^{+}} [\ln|x|]_{a}^{3} = \lim_{a \to0^{+}} (\ln|3| - \ln|a|)

STEP 5

As aa approaches0 from the right, lna\ln|a| approaches negative infinity. Therefore, the expression ln3lna\ln|3| - \ln|a| approaches infinity.
So, the integral 031xdx\int_{0}^{3} \frac{1}{x} d x diverges.

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