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

Calculate βˆ«π‘’^(π‘₯) (1 βˆ’ 𝑒^(βˆ’π‘₯))Β² dπ‘₯.

04:03

Video Transcript

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

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

However, because we’re only dealing with multiplication, we can also multiply 𝑒 to the π‘₯ power times one minus 𝑒 to the negative π‘₯ power. And then from there, we’ll take whatever we get and multiply that by one minus 𝑒 to the negative π‘₯ power. Let’s do the second option.

We’ll distribute this 𝑒 to the π‘₯ power over one minus 𝑒 to the negative π‘₯, which is 𝑒 to the π‘₯, minus 𝑒 to the π‘₯ power times 𝑒 to the negative π‘₯ power. We have to be careful here. Remembering our power rules, if we’re multiplying π‘Ž to the π‘š power times π‘Ž to the 𝑛 power, what we do is add the exponents, π‘š plus 𝑛. In this case, that means we’ll be adding π‘₯ plus negative π‘₯. 𝑒 to the π‘₯ power times 𝑒 to the negative π‘₯ power equals 𝑒 to the zero power, which equals one.

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

What we have here 𝑒 to the π‘₯ minus two plus 𝑒 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 𝑒 to the π‘₯ power plus 𝑒 to the negative π‘₯ power minus two dπ‘₯. And this is much more manageable. We’ll just find the integral of each individual terms and add them together. The integral of 𝑒 to the π‘₯ dπ‘₯ equals 𝑒 to the π‘₯ power.

The integral of 𝑒 to the negative π‘₯ power with respect to π‘₯ is negative 𝑒 to the negative π‘₯ power. And we’re subtracting the integral of two with respect to π‘₯, which is negative two π‘₯ plus any constant 𝑐. The integral is 𝑒 to the π‘₯ power minus 𝑒 to the negative π‘₯ power minus two π‘₯ plus 𝑐. The key here was to simplify the expression that we were originally given into a format that was easier to integrate.

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