The integrals in Exercises 1–34 converge. Evaluate the integra...

Question

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.

∫₋∞^∞ (x dx) / (x² + 4)^(3/2)

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the improper integral. The integral evaluated from negative infinity to positive infinity of x divided by parentheses of x 2 + 9 of 3 divided by 2 dx. OK. So it appears for this particular prompt, we're ultimately asked to evaluate this improper integral that is given to us by the prom itself. So now that we know that we're ultimately solving for this improper interval, or rather we're evaluating this improper integral, we need to recall and note that in order to evaluate this specific integral, we will first need to split it into two separate integrals at x is equal to 0, and we will need to use limits. So we need to go ahead and write the integral evaluated from negative infinity to positive infinity of x divided by parentheses of x 2 + 9 to the power of 3 divided by 2. D x minus the integral evaluated from negative infinity to 0 of x divided by parentheses x to the 2 + 9 in parentheses to the power of 3 divided by 2. Plus, the integral evaluated from 0 to positive infinity of x divided by parentheses x 2 + 9 to the power of 3 divided by 2 dx, and don't forget that each of our integrals need to have a dx after it. After our expression. OK. So, our first step, so this is our first major step towards solving this problem, is we need to rewrite the integral with a limit. So we need to take the integral evaluated from negative infinity to 0 x divided by x 2 + 9 in parentheses to the power of 3 divided by 2. D X. Plus the integral evaluated from 0 to positive infinity of x divided by parentheses x to the power of 2. Plus 9 in parentheses to the power of 3 divided by 2. D X, which will be all equal to the limit as A approaches negative infinity of the integral evaluated from A to 0 of X divided by parentheses. X to the power of 2. Plus 9, and don't forget that this is in parentheses to the power of 3 divided by 2, and don't forget your DX. Plus the limit as B approaches positive infinity of the integral evaluated from 0 to B of x divided by parentheses x to the 2 + 9 in parentheses to the power of 3 halves d x. Fantastic. Our second step now is we need to find the antiderivative of, so once again we're finding the antiderivative of the integral of x divided by parentheses x 2, plus 9 in parentheses to the power of 3 divided by 2 dx. And to do this, we need to recall and use the substitution method. So, let us state that U for this particular prompt will be equal to x 2 + 9. So that will mean that DU by DX will be equal to 2x, which will mean that 1/2 multiplied by DU will be equal to x multiplied by DX. So that means we can go ahead and write that the integral of x divided by parentheses x 2 + 9 in parentheses the power of 3 halves, the x will be equal to 1/2 multiplied by the integral of 1 divided by u to the power of 3 divided by 2. D u, which will be equal to 1/2 multiplied by the integral of u to the power of -3 divided by 2. Fantastic duu, which will be equal to 1/2 multiplied by u to the power of -3 divided by 2 + 1 divided by 3 divided by 2 + 1 + c where c denotes some constant variable, and this will be equal to 1/2. Multiplied by u to the power of -1/2. Divided by -1/22. Plus C and we need to keep going. Yes, we need to make sure that we have it in its most simple form, so we need to keep simplifying. So, this will be equal to -1 divided by the square root of U + C, which will be equal to -1 divided by the square root of X2 + 9. Plus C, fantastic. So, moving right along here, our third step, this is step 3, is we need to evaluate our definite integral. So we need to take the integral evaluated from A to 0 of X divided by parentheses of x 2 + 9 in parentheses to the power of 3 divided by 2. The X, which will be equal to -1 divided by the square root of x 2 + 9, and this expression will be evaluated from A to 0. And this will be equal to -1 divided by the square root of 02 + 9. Minus parentheses of 1 divided by the square root of a 2 + 9. Which will be equal to 1/3 + 1 divided by the square root of A2 + 9, fantastic. And we can also go ahead and write that the integral evaluated from 0 to b of x divided by parentheses x 2 plus 9 in parentheses to the power of 3 divided by 2. The x will be equal to 1 divided by the square root of x 2 plus 9, and this expression will be evaluated from 0. To be And this will be equal to -1 divided by the square root of B 2 + 9 minus. Parentheses 1 divided by the square root of 02 + 9. Fantastic, and this will be equal to -1 divided by the square root of b to the 2 + 9. Plus 1 divided by 3. Fantastic, when we write it in its most simplified form. So our 4th step is we need, so once again, this is step number 4, and our 4th step is, is we need to take the limit as A approaches, so once again we're taking the limit as A approaches negative infinity. So when we take the limit as a approaches negative infinity, we can go ahead and write that the limit as a approaches infinity of parentheses of 1/3 plus 1 divided by the square root of a 2 + 9. And this will be equal to 1/3 + 0, which will be equal to -13. Fantastic, moving right along here. So that will mean that our fifth step, so this is step number 5, is we need to take the limit as B approaches positive infinity. So that means we need to write that the limit as B approaches positive infinity of parentheses 1 divided by the square root of b 2 + 9 + 1/3 will be equal to 0 + 1/3, which will be equal to 1/3 simply. So for step number 6, which is our final step, we need to combine both of our limits. So we need to take the limit as A approaches negative infinity of the integral evaluated from a to 0 of x divided by parentheses, x to the 2 + 9 in parentheses to the power of 3 divided by 2. D x plus the limit as b approaches positive infinity of the integral evaluated from 0 to b of x divided by parentheses of x to the 2 + 9 in parentheses to the power of 3 divided by 2 d x. And this will be equal to 1/3 plus 1/3, which will be simply equal to 0, and 0 is our correct and final answer. All righty, we did it. So going back to the top to look at the problem itself and to recap what we've learned. We can safely say that our final answer for our evaluated improper integral once again is, without a doubt, simply 0, and that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.∫₋∞^∞ (x dx) / (x² + 4)^(3/2)

Watch next

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₋∞^∞ (x dx) / (x² + 4)^(3/2)