Video: APCALC03AB-P1A-Q28-854185437979

Which of the following is equivalent to the integral from βˆ’1 to 3 of 𝑓(π‘₯ βˆ’ 2) dπ‘₯? [A] The integral from 1 to 5 of 𝑓(π‘₯) dπ‘₯. [B] The integral from 1 to 5 of 𝑓(π‘₯ + 2) dπ‘₯. [C] The integral from βˆ’3 to 1 of 𝑓(π‘₯ + 2) dπ‘₯. [D] the integral from βˆ’3 to 1 of 𝑓(π‘₯) dπ‘₯.

03:30

Video Transcript

Which of the following is equivalent to the integral from negative one to three of 𝑓 of π‘₯ minus two with respect to π‘₯. a) The integral from one to five of 𝑓 of π‘₯ with respect to π‘₯. b) The integral from one to five of 𝑓 of π‘₯ plus two with respect to π‘₯. c) The integral from negative three to one of 𝑓 of π‘₯ plus two with respect to π‘₯. Or d) the integral from negative three to one of 𝑓 of π‘₯ with respect to π‘₯.

Now, letβ€²s consider what has happened here. We can see that the limits in our original integral are different from the limits in each of the options weβ€²ve been given. We can also see that although weβ€²re integrating with respect to the same variable, we have dπ‘₯. The argument of the function has changed. We had π‘₯ minus two initially. And we now have either π‘₯ or π‘₯ plus two in the options weβ€²ve been given.

We can answer this question by considering transformations of graphs. We donβ€²t know what our function 𝑓 of π‘₯ looks like. But letβ€²s say for the sake of argument it looks something like this. 𝑓 of π‘₯ minus two is a translation of the graph of 𝑓 of π‘₯ by the vector two, zero. Itβ€²s a horizontal translation, two units in the positive π‘₯-direction. So perhaps 𝑓 of π‘₯ minus two looks something like this. Now, the integral weβ€²ve been given to find the area under the curve of 𝑓 of π‘₯ minus two between the limits of negative one and three. Thus perhaps this area here shaded in orange.

Now, as weβ€²ve just translated the original graph, this area will be equivalent to an area below the original graph, but with different limits. Now, the graph of 𝑓 of π‘₯ is two units to the left of the graph of 𝑓 of π‘₯ minus two. And so the limits for the pink area will be two values less than the limit for the orange area. That will be negative three for the lower limit and one for the upper limit. The pink area, which remember is equivalent to the orange area, our original integral, is therefore equal to the integral from negative three to one of 𝑓 of π‘₯ with respect to π‘₯. Thatβ€²s option d in the four options we weβ€²ve given.

We could also show this using the method of substitution if we wished. We could define a new variable, 𝑒, which is equal to π‘₯ minus two. d𝑒 by dπ‘₯ is equal to one. And therefore, d𝑒 is equal to dπ‘₯. When π‘₯ equals three, 𝑒 would be equal to three minus two, which is one. And when π‘₯ equals negative one, 𝑒 equals negative one minus two, which is negative three. Our integral in terms of 𝑒 then would become the integral from negative three to one of 𝑓 of 𝑒 d𝑒. But as this is a definite integral, it doesnβ€²t matter what variable we use. It would be equivalent if we used 𝑑 or 𝑝. Or we could return to π‘₯. Because weβ€²re going to substitute the values of our limits once we perform the integral, the 𝑒s or π‘₯s or whatever variable weβ€²re integrating with respect to will be replaced. So it doesnβ€²t matter what variable we use.

We see again then that the integral which is equivalent to the integral from negative one to three of 𝑓 of π‘₯ minus two with respect π‘₯ is option d. The integral from negative three to one of 𝑓 of π‘₯ with respect to π‘₯.

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