# Video: AP Calculus AB Exam 1 β’ Section I β’ Part A β’ Question 6

Calculate β«π^(π₯) (1 β π^(βπ₯))Β² dπ₯.

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### Video Transcript

Calculate the integral of π to the π₯ power times one minus π to the negative π₯ power squared dπ₯.

The first thing that we can do here is take this one minus π to the negative π₯ power squared and expand it. Weβre saying one minus π to the negative π₯ power times itself. One minus π to the negative π₯ power. Bring down the π to the π₯ and the dπ₯. Inside this integral, weβre multiplying these three terms together. To simplify, we could multiply one minus π to the negative π₯ times one minus π to the negative π₯. And then, whatever we got there, we would multiply by π to the π₯ power.

However, because weβre only dealing with multiplication, we can also multiply π to the π₯ power times one minus π to the negative π₯ power. And then from there, weβll take whatever we get and multiply that by one minus π to the negative π₯ power. Letβs do the second option.

Weβll distribute this π to the π₯ power over one minus π to the negative π₯, which is π to the π₯, minus π to the π₯ power times π to the negative π₯ power. We have to be careful here. Remembering our power rules, if weβre multiplying π to the π power times π to the π power, what we do is add the exponents, π plus π. In this case, that means weβll be adding π₯ plus negative π₯. π to the π₯ power times π to the negative π₯ power equals π to the zero power, which equals one.

We need to multiply π to the π₯ power minus one times one minus π to the negative π₯ power. π to the π₯ power times one equals π to the π₯. And now, we have π to the π₯ power times negative π to the negative π₯ power. So it looks like this: minus π to the power times π to the negative π₯ power, which we know is π to the zero power. Then, we have negative one times one which is negative one and negative one times negative π to the negative π₯ power which is positive π to the negative π₯ power. Bringing down what we know, π to the π₯ power minus π to the zero β which is one β minus one plus π to the negative π₯ power. One minus one equals negative two.

What we have here π to the π₯ minus two plus π to the negative π₯ is a simplified expression of what we started with. We can rewrite our original integral to say we wanna find the integral of π to the π₯ power plus π to the negative π₯ power minus two dπ₯. And this is much more manageable. Weβll just find the integral of each individual terms and add them together. The integral of π to the π₯ dπ₯ equals π to the π₯ power.

The integral of π to the negative π₯ power with respect to π₯ is negative π to the negative π₯ power. And weβre subtracting the integral of two with respect to π₯, which is negative two π₯ plus any constant π. The integral is π to the π₯ power minus π to the negative π₯ power minus two π₯ plus π. The key here was to simplify the expression that we were originally given into a format that was easier to integrate.