# Video: MATH-DIFF-INT-2018-S1-Q14

If β«_(β2) ^(3) π(π₯)dπ₯ = 12 and β«_(β2) ^(5) π(π₯)dπ₯ = 16, find β«_(3) ^(5) π(π₯)dπ₯.

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### Video Transcript

If the integral of π of π₯ dπ₯ between the limits of negative two and three is equal to 12 and the integral of π of π₯ dπ₯ between the limits of negative two and five is equal to 16, find the integral of π of π₯ dπ₯ between the limits of three and five.

To begin this question, let us consider the graph of some function π of π₯ in the π₯π¦-plane. Now, we know that one interpretation for the value of an integral is the area bounded by the curve, the π₯-axis, and the limits of the integral. If we take this integral as an example between the limits of negative two and five, the value of this integral will then be the bounded region shown on our diagram between the curve, the π₯-axis, and the lines π₯ equals minus two and π₯ equals five.

By similar reasoning, if we were instead to look at the integral between negative two and three, this would be the area bounded by the curve, the π₯-axis, and the lines π₯ equals negative two and π₯ equals three, as shown on the diagram. Finally, the area that weβre interested in is bounded by our curve, the π₯-axis, and the lines π₯ equals three and π₯ equals five. Looking at our diagram, we should be able to see that this area is the difference between the previous two areas that we outlined. Now, here, you may recall that this diagram uses an example curve for π of π₯.

What if instead the real curve past below the π₯-axis? You may recall that for sections where the curve passes below the π₯-axis, integrals will evaluate to negative numbers which gives negative areas. This would be a valid concern for our question if we were indeed dealing with the area under a curve. In our case, however, weβre only worried about the value of the integrals. And therefore, we donβt need to worry about splitting things up to get rid of negative areas. Okay, this is a nice illustration. But how do we write things more formally?

Firstly, letβs take our integral between the limits of negative two and five. For this question, these limits represent the entire range over which weβre interested in. Now, one of the properties of integrals is that we can split our integral into component parts as long as we respect the original upper and lower bounds of five and negative two. Here, youβll notice weβve used π for the lower bound of our first integral and π for the upper bound of our second integral. So in a sense, we havenβt created any gaps over the range. From this equation, weβre saying that the value of our original integral is equal to the sum of its component parts. Now, we can begin to see how this helps us with the question.

If we were to take the case where π is equal to three, by doing this, we now have an equation which contains the three integrals present in the question. If we now subtract the integral of π of π₯ dπ₯ between the limits of negative two and three from both sides of the equation, we reach the same conclusion that we did with our example diagram. The value of the integral that weβre interested in is equal to the value of the integral between the limits of negative two and five subtract the value of the integral between the limits of negative two and three. Another way of saying this is itβs the difference between the value of these two integrals.

Now that we understand whatβs going on, this question reduces to a very easy sum. We can substitute in the values given in the question, which are 16 for our first integral and 12 for our second integral. 16 minus 12 is just four. And therefore, we found our answer. The value of the integral of π of π₯ dπ₯ between the limits of three and five is equal to four.