Math  /  Calculus

QuestionNote: You may need to assume the fact that limM+MneM=0\lim_{M \to +\infty} M^n e^{-M} = 0 for all nn.
Decide whether or not the given integral converges.
81x2dx\int_{-\infty}^{-8} \frac{1}{x^2} dx
The integral converges. The integral diverges.
If the integral converges, compute its value. (If the integral diverges, enter DNE.)

Studdy Solution

STEP 1

What is this asking? Does this integral have a finite value, and if so, what is it? Watch out! The integral goes to negative infinity, so we need to be careful with our signs!

STEP 2

1. Rewrite as a limit
2. Find the indefinite integral
3. Evaluate the limit

STEP 3

Let's **rewrite** this improper integral as a limit!
We replace the negative infinity with a variable, let's call it tt, and then take the limit as tt approaches negative infinity.
This gives us: limtt81x2dx \lim_{t \to -\infty} \int_{t}^{-8} \frac{1}{x^2} dx This is how we handle integrals that go off to infinity!

STEP 4

Now, let's **find** the indefinite integral of 1x2\frac{1}{x^2}.
Remember, 1x2\frac{1}{x^2} is the same as x2x^{-2}.
Using the power rule for integration, we add 1 to the exponent and then divide by the new exponent. x2dx=x2+12+1=x11=1x \int x^{-2} dx = \frac{x^{-2+1}}{-2+1} = \frac{x^{-1}}{-1} = -\frac{1}{x} Don't forget to add the constant of integration, CC, when finding an indefinite integral.
However, since we're working with a definite integral, the constant will disappear when we evaluate the bounds, so we can skip it here.

STEP 5

Now, let's **evaluate** the definite integral from tt to 8-8: [1x]t8=18(1t)=18+1t \left[ -\frac{1}{x} \right]_t^{-8} = -\frac{1}{-8} - \left(-\frac{1}{t}\right) = \frac{1}{8} + \frac{1}{t}

STEP 6

Finally, let's **take** the limit as tt approaches negative infinity: limt(18+1t) \lim_{t \to -\infty} \left(\frac{1}{8} + \frac{1}{t}\right) As tt gets super-duper negative, 1t\frac{1}{t} gets incredibly close to zero.
So, we have: limt(18+1t)=18+0=18 \lim_{t \to -\infty} \left(\frac{1}{8} + \frac{1}{t}\right) = \frac{1}{8} + 0 = \frac{1}{8}

STEP 7

The integral converges, and its value is 18\frac{1}{8}.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord