Paul's Online Notes
Home / Calculus I / Integrals / Computing Definite Integrals
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 5.7 : Computing Definite Integrals

9. Evaluate the following integral, if possible. If it is not possible clearly explain why it is not possible to evaluate the integral.

$\int_{2}^{1}{{\frac{{2{y^3} - 6{y^2}}}{{{y^2}}}\,dy}}$

Show All Steps Hide All Steps

Start Solution

In this case we’ll first need to simplify the integrand to remove the quotient before we actually do the integration. Doing that integrating the function gives,

$\int_{2}^{1}{{\frac{{2{y^3} - 6{y^2}}}{{{y^2}}}\,dy}} = \int_{2}^{1}{{2y - 6\,dy}} = \left. {\left( {{y^2} - 6y} \right)} \right|_2^1$

Do not get excited about the fact that the lower limit of integration is larger than the upper limit of integration. This will happen on occasion and the integral works in exactly the same manner as we’ve been doing them.

Also, recall that we don’t need to add the “+c” in the definite integral case as it will just cancel in the next step.

Show Step 2

The final step is then just to do the evaluation.

We’ll leave the basic arithmetic to you to verify and only show the results of the evaluation. Make sure that you evaluate the upper limit first and then subtract off the evaluation at the lower limit.

Here is the answer for this problem.

$\int_{2}^{1}{{\frac{{2{y^3} - 6{y^2}}}{{{y^2}}}\,dy}} = - 5 - \left( { - 8} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{3}$