Integrals 215 problem 11.mws

Integration by parts is just the counterpart in integration theory of the product rule in differentiation theory. One way to do integration by parts is to "guess" an anti-derivative of your function by finding some product of functions such that by differentiating using the product rule, one of the terms becomes exactly your function, while if you are lucky, the other term becomes something you can already find an anti-derivative of. For example, suppose that you want to find the anti-derivative of , you first think of some product of two functions such that when you differentiate, one of the terms becomes . About the only thing that we can guess is . Now when you differentiate, you get

> diff(x*ln(x),x);

This is not exactly what we want, but it is easy to think of a function whose derivative is -1.

> diff(-x,x);

Adding these two functions gives us the desired anti-derivative. In other words,

> diff(x*ln(x)-x,x);

One nice thing about this method of integration by parts is that we explicitly construct an anti-derivative of the function.  Thus if we are computing a definite integral, we can simply plug in the limits of integration to obtain the answer.  Sometimes there is more than one natural guess for the product, and you may have to use more than one guess to get the answer. For example, if , I can easily think of two ways to write products of two functions such that the derivative of one of the terms is f(x). First, we could take , because

> diff(-x*cos(ln(x)),x);

The problem is that while we get one term we like, the second term is just as nasty as the original. Another guess for the anti-derivative is , which when we differentiate gives

> diff(x*sin(ln(x)),x);

This is also not so great, but if you look at both answers, you can find the anti-derivative you want.

Submission:

(a) Determine an anti-derivative of f(x)=sin(ln(x)) from the information above.

(b) Guess a function which is a product such that when you differentiate this product, one of the terms in the derivative is . Use that guess to explicitly find an anti-derivative of . Then compute in Maple to see if you have the correct answer.

(c) Guess a function which is a product such that when you differentiate this product, one of the terms in the derivative is . You probably won't find that the second term is any easier to integrate than the original. So try again with a different function. Then put the two answers together to find an anti-derivative of . Check your answer in Maple.

Submission worksheet:

The standard method of introducing integration by parts is to consider an integrand as a product of two terms, one an expression , and the other a differential , so that We then differentiate the function u and integrate dv to obtain the rule for integration by parts:

Of course, we can use Maple to help with both the integrals and the derivatives involved. For example, let us consider the integral from the previous activity. To integrate by parts, we need to identify a u and a dv . The principle is that the dv should be something reasonable to integrate, while the u can be more complicated, because we can always differentiate a function. In this case it is very natural to decompose the integrand as follows.

> u:=ln(x); dv:=1;

Then we differentiate the first expression and integrate the second one to obtain:

> du:=diff(ln(x),x); v:=int(dv,x);

Thus according to our principle of integration by parts, we obtain

> Int(u*dv,x)=u*v-Int(v*du,x);

Since the last integral can be done very easily, we obtain the result

> Int(u*dv,x)=u*v-int(v*du,x);

Notice that this is exactly the anti-derivative we obtained in the discussion in the first activity.  Consider the integral from the first activity. Let us split the integrand up in the only reasonable manner that we can.

> u:=sin(ln(x));dv:=1;

> du:=diff(u,x);v:=int(dv,x);

> Int(u*dv,x)=u*v-Int(v*du,x);

We run into a similar problem as we did before, because the integral is equally complicated as the first integral. But we proceed to investigate how to integrate this function by parts, by decomposing it as a product.

> u:=cos(ln(x));dv:=1;

> du:=diff(u,x);v:=int(dv,x);

> Int(u*dv,x)=u*v-Int(v*du,x);

Now, looking at both expressions, we can see how to do the original integral.

Submission:

(a) Explain how to use the two integral calculations above to find the indefinite integral .

(b) Break the integrand in up into two parts using the method above, and then compute the integral. Check your result by computing the integral in Maple directly.

(c) Use the method above to find the indefinite integral .

Submission worksheet:

Consider the indefinite integral .

Submission:

(a) Use integration by parts to find the indefinite integral by hand.

(b) Find the indefinite integral using Maple.

(c) Illustrate, and check that your answer is reasonable, by graphing both the function and its antiderivative (Let C=0).  Use complete sentences to explain why your answer is reasonable. Pay attention to the "five c's:" clarity, conciseness, completeness, cohesiveness and correctness.

Submission worksheet: