# Video: AP Calculus AB Exam 1 β’ Section I β’ Part A β’ Question 16

If π is continuous for all real numbers π₯ and β«ββ΄ π(π₯) dπ₯ = 20, calculate β«ββΒ³ [π(π₯ + 1) β 3π₯ + 1] dπ₯.

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### Video Transcript

If π is continuous for all real numbers π₯ and the integral from zero to four of ππ₯ with respect to π₯ equals 20, calculate the integral from negative one to three of π of π₯ plus one minus three π₯ plus one with respect to π₯.

For this question, weβve been given the value of an integral which involves some function π. And weβve then been asked to find the value of another integral which involves the same function π. However, this time it operates on π₯ plus one instead of π₯. Before we consider how to deal with this, let us first split our second integral into two parts using the laws of integration. If we were to take the integral of π of π₯ plus one with respect to π₯ and add it to the integral of negative three π₯ plus one with respect to π₯ both between the intervals of negative one and three, this would be equivalent to the integral given in our question.

Now, letβs go back to considering our π of π₯ and our π of π₯ plus one. Letβs imagine that the graph of our function π of π₯ looks like this. If we say that π¦ is equal to π of π₯, we can conceptualize that the value of our first integral is represented by the area under the curve. And this is bounded by the curve itself, the π₯-axis, and the limits of our integral itself, which are the lines π₯ equals zero and π₯ equals four.

Now, letβs consider a general shift in our function π of π₯. We know that π of π₯ minus π would correspond to a horizontal shift of the curve shown here of π units in the positive π₯-direction. Since weβre looking at π of π₯ plus one, this would then correspond to a horizontal shift of negative one units in the positive π₯-direction or a shift of one unit to the left. At this point, let us look back to the original integral sum that we created. We may notice that the limits of our π of π₯ plus one integral have also been shifted negative one units in the π₯-direction when compared to our π of π₯ integral.

If we now look back at what this represents on our graph, we may notice that since the curve and the limits have all been shifted negative one units in the positive π₯-direction, the area that weβve marked under our π of π₯ plus one curve is the same as the original area that we marked under our π of π₯ curve. Again, we should reiterate that this is only true because the limits of our integral have shifted to match the shift in our curve. Through this process, we can logically conclude that the integral between negative one and three of π of π₯ plus one with respect to π₯ is the same as the integral between zero and four of π of π₯ with respect to π₯.

Now, a more rigorous way to mathematically show this would be to do a π’ substitution, replacing our π₯ plus one, our dπ₯, and the limits of our integration in the following way. However, in this case, the substitution would be fairly trivial. Nonetheless, weβve shown here how you arrive at the same result. Now looking back at our integral sum, we note that we already have a value for our first integral. And this has been given by the question. The only steps that we have left are therefore to evaluate our second integral. Integrating negative three π₯ plus one, we get negative three over two π₯ squared plus π₯.

We now substitute in the limits of our integration and we evaluate. If we continue to simplify, we eventually arrive at 20 minus 16 over two which is of course just 20 minus eight. Hence, we arrive at our final answer which is 12. We have now answered our question. And we have calculated that the given integral is equal to 12.

We arrived at this answer first by splitting our integral into a sum of two separate integrals, next by recognising the implications in the shift of our function and the shift of the integral limits, and finally by evaluating and simplifying the two parts of the integral sum that we created.