Math

QuestionDetermine if the integral 141x2/3dx\int_{-1}^{4} \frac{1}{x^{2 / 3}} d x converges or diverges.

Studdy Solution

STEP 1

Assumptions1. We are asked to evaluate the improper integral 141x/3dx\int_{-1}^{4} \frac{1}{x^{ /3}} dx. . We need to determine whether this integral converges or diverges.

STEP 2

The given integral is an improper integral because the function f(x)=1x2/f(x) = \frac{1}{x^{2 /}} is not defined at x=0x =0. So, we need to split the integral at x=0x =0 and evaluate each part separately.
141x2/dx=101x2/dx+041x2/dx\int_{-1}^{4} \frac{1}{x^{2 /}} dx = \int_{-1}^{0} \frac{1}{x^{2 /}} dx + \int_{0}^{4} \frac{1}{x^{2 /}} dx

STEP 3

Let's evaluate the first integral 101x2/3dx\int_{-1}^{0} \frac{1}{x^{2 /3}} dx.
We can rewrite the integral as a limit101x2/3dx=lima01a1x2/3dx\int_{-1}^{0} \frac{1}{x^{2 /3}} dx = \lim_{a \to0^-} \int_{-1}^{a} \frac{1}{x^{2 /3}} dx

STEP 4

Now, we can evaluate the integral inside the limit.
lima01a1x2/3dx=lima0[3x1/3]1a\lim_{a \to0^-} \int_{-1}^{a} \frac{1}{x^{2 /3}} dx = \lim_{a \to0^-} \left[3x^{1 /3}\right]_{-1}^{a}

STEP 5

Evaluate the limit.
lima0[3a1/33(1)1/3]=\lim_{a \to0^-} \left[3a^{1 /3} -3(-1)^{1 /3}\right] = -\inftyThe first integral diverges to -\infty.

STEP 6

Now, let's evaluate the second integral 041x2/3dx\int_{0}^{4} \frac{1}{x^{2 /3}} dx.
We can rewrite the integral as a limit041x2/3dx=limb0+b41x2/3dx\int_{0}^{4} \frac{1}{x^{2 /3}} dx = \lim_{b \to0^+} \int_{b}^{4} \frac{1}{x^{2 /3}} dx

STEP 7

Now, we can evaluate the integral inside the limit.
limb0+b41x2/3dx=limb0+[3x1/3]b4\lim_{b \to0^+} \int_{b}^{4} \frac{1}{x^{2 /3}} dx = \lim_{b \to0^+} \left[3x^{1 /3}\right]_{b}^{4}

STEP 8

Evaluate the limit.
limb0+[3(4)1/33b1/3]=\lim_{b \to0^+} \left[3(4)^{1 /3} -3b^{1 /3}\right] = \inftyThe second integral diverges to \infty.

STEP 9

Since both the integrals diverge, the original integral also diverges.
So, the integral 4x2/3dx\int_{-}^{4} \frac{}{x^{2 /3}} dx diverges.

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