Let ( f ) be a continuous function with domain ( mathbbR...

Let ( f ) be a continuous function with domain ( mathbb{R} ). We define ( F(x)=int_{0}^{x} f(t) d t ) Given the following table: egin{tabular}{|l|l|l|} hline( x ) & ( f(x) ) & ( F(x) ) \ hline 1 & 4 & 3 \ hline 2 & 1 & 4 \ hline 3 & 2 & 3 \ hline 4 & 3 & 1 \ hline 5 & 4 & 2 \ hline end{tabular} Compute ( int_{1}^{2} x fleft(x^{2}+1 ight) d x ). Hint: use u-substitution method and FTC part 2. Enter your answer in decimal form. Round to two decimal places if needed (e.g. enter ( 0.1 ) as ( 0.1 ), enter ( 0.2345 ) as ( 0.23 ) or ( 0.24 ) ).

Transcript

00:01 If we want to evaluate this integral here, it gives us a hint to start by using a u substitution.

00:12 Now, where i would think would be a good idea is to take the most complicated part of this and then choose that to be our u.

00:23 And in this case, i would say what's on the inside of our function, this x squared plus 1, is kind of complicating everything.

00:32 Because if this was just like x, then we could use like integration by parts.

00:43 I don't know why i blanked on the name to get this pretty straightforward.

00:49 However, since we have this here, i really don't know how to tackle it right away.

00:55 So what we can do instead is just let this be you and then plug everything in.

01:02 So first, we're going to have u is equal to x squared plus 1.

01:07 Now we need to take the derivative of this with respect x.

01:10 We get du or dx is equal to.

01:13 So we use power rule.

01:14 So 2x and then that's just zero.

01:16 And then to get our differential, we kind of move the dx over, but it's a little bit more than that.

01:23 But we get du is equal to 2x.

01:26 Dx.

01:31 So if we were to come over here and turn this x into a 2x, then that would just be equal to du.

01:41 But we just can't multiply by two.

01:44 So i'm going to multiply by one -half as well.

01:48 So now these terms just turn into d -u.

01:53 So it would be one -half integral, and i'm going to not touch these bounds for right now.

02:00 But that is going to give f of u, and then 2x -d -x, as we showed over here, just turns into d -u.

02:18 Now, we need to change our bounds.

02:26 So, when x is equal to, so our lower bound is 1, then u is going to be, so it is 1 squared plus 1, which is just going to be 2.

02:40 So our lower bound now becomes 2, and when x is equal to 2, then u is going to become, so it would be 2.

02:56 Squared plus 1, so 4 plus 1, 5...

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