Video Transcript
Approximate the integral from zero to one of the sin of 𝑥 squared with respect to 𝑥 using the first two terms of an appropriate series.
In this question, we’re given a definite integral. We need to approximate the value of this definite integral. And to do this, we’re told to use the first two terms of an appropriate series. The first thing we can think of is if we were able to approximate our integrand the sin of 𝑥 squared, then we could integrate our approximation instead. And this might be easier. But we don’t know an approximation for the sin of 𝑥 squared. We do, however, know an approximation for the sin of 𝑥.
We recall the Maclaurin series for the sin of 𝑥. It’s equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power divided by two 𝑛 plus one factorial multiplied by 𝑥 to the power of two 𝑛 plus one. And this converges for all real values of 𝑥. This is a useful series which we should commit to memory. But we don’t want an approximation for the sin of 𝑥. We want one for the sin of 𝑥 squared. Well, we have this expression for the sin of 𝑥. We could just write 𝑥 squared for every instance of 𝑥. And then we would have an expression for the sin of 𝑥 squared. And because our series converges for all real values of 𝑥, our new series will also converge for all real values of 𝑥.
So this gives us the sin of 𝑥 squared is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power divided by two 𝑛 plus one factorial times 𝑥 squared raised to the power of two 𝑛 plus one for all real values of 𝑥. And now we can use this to approximate our integral. We’ll only use the first two terms of this series. So this gives us the sin of 𝑥 squared is approximately equal to the sum from 𝑛 equals zero to one of negative one to the 𝑛th power divided by two 𝑛 plus one factorial times 𝑥 squared to the power of two 𝑛 plus one.
Now, we just need to expand this series. We’ll start when 𝑛 is equal to zero. Substituting 𝑛 is equal to zero into our summand, we get negative one raised to the zeroth power divided by two times zero plus one factorial multiplied by 𝑥 squared raised to the power of two times zero plus one. And of course, we can simplify this. First, negative one raised to the zeroth power is equal to one. Next, two times zero plus one is equal to one. But then one factorial is just equal to one. So our coefficient just simplifies to give us one divided by one, which is, of course, just equal to one.
Next, two times zero plus one is, of course, just equal to one. And any number raised to the power of one is just equal to itself. So our first term just simplifies to give us 𝑥 squared. We’ll do exactly the same to find the second term in our series. That’s when 𝑛 is equal to one. Substituting 𝑛 is equal to one into our summand, we get negative one raised to the first power divided by two times one plus one factorial multiplied by 𝑥 squared raised to the power of two times one plus one.
And once again, we can simplify. First, negative one raised to the first power is just equal to negative one. Next, two times one plus one is equal to three, and then three factorial is equal to three times two times one, which is, of course, equal to six. Now, we want to find our exponent of 𝑥. First, two times one plus one is equal to three. And to find our exponent of 𝑥 in this situation, we’re going to need to use our laws of exponents. 𝑎 raised to the power of 𝑏 all raised to the power of 𝑐 is equal to 𝑎 raised to the power of 𝑏 times 𝑐.
In our case, our value of 𝑎 is 𝑥, our value of 𝑏 is two, and our value of 𝑐 is three. So this is equal to 𝑥 to the power of two times three, which is 𝑥 to the sixth power. So what we’ve shown is the sin of 𝑥 squared is approximately equal to 𝑥 squared minus 𝑥 to the sixth power divided by six. And we can now use this approximation to approximate the integral from zero to one of the sin of 𝑥 squared with respect to 𝑥. Since the sin of 𝑥 squared is approximately equal to 𝑥 squared minus 𝑥 to the sixth power over six, we have the integral from zero to one of the sin of 𝑥 squared with respect to 𝑥 is approximately equal to the integral from zero to one of 𝑥 squared minus 𝑥 to the sixth power over six with respect to 𝑥.
But this integral is much easier to evaluate. We can just do this by using the power rule for integration. We need to add one to our exponents of 𝑥 and then divide by this new exponent. This gives us 𝑥 cubed over three minus 𝑥 to the seventh power divided by seven times six evaluated at limits of integration zero and one. Now, all we need to do is evaluate this at the limits of integration. However, we see if we substitute zero into this expression, we just get zero. So we only need to substitute 𝑥 is equal to one. This gives us one cubed divided by three minus one to the seventh power divided by 42.
And if we calculate this expression, we get 13 divided by 42. Therefore, by using the Maclaurin series for the sin of 𝑥, we were able to find a power series representation for the sin of 𝑥 squared. We were then able to use the first two terms of this power series representation to approximate the value of the integral from zero to one of the sin of 𝑥 squared with respect to 𝑥. We got that this integral was approximately equal to 13 divided by 42.
