Integral Equations and Differentiation under the Integral Sign

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Take the integral from zero to one of x cubed minus one, over the natural log of x, d x. Try substitution, try by parts. Nothing works, right? None of our usual methods gives an antiderivative here. And yet, the answer is simply the natural log of four. So how is that possible? And one more puzzle: what if the function you want is hidden inside its own integral? Today: integral equations, and differentiation under the integral sign. Both come up in JEE problems. So, let's begin.

0:40Type 1: x sits in a limit

So, the first type. On screen: f of x equals g of x, plus the integral from a to x of k of t times f of t, d t. Here g and k are known, and f is the unknown. The long S is the integral sign, a and x are its limits, and t is a dummy letter. See the difficulty? The unknown f sits inside the integral. Okay? So tell me, how do we get rid of that integral sign? We differentiate, yes, exactly! From the previous concept: for a continuous integrand, the derivative of the integral from a to x is the integrand at x, that is, k of x times f of x. The small dash is a prime: f prime means the derivative of f. So f prime of x equals g prime of x, plus k of x times f of x. That is a differential equation. Right? But a differential equation also needs a starting value. Look at the first line again. What is f of a? Put x equal to a. The integral from a to a, of whatever is inside, is zero. The dots stand for the same integrand. So f of a equals g of a. Our starting value, for free. Clear? Let us try a course example. f of x equals one, plus the integral from zero to x of t times f of t, d t. Here g is one, and k of t is t. Differentiate: f prime equals x times f. The double arrow means, so it follows. And at x equal to zero, f of zero is one. With me? Now solve it. Divide by f: f prime over f equals x. Integrate both sides. On the left, the natural log of mod f, where the straight bars are the modulus. On the right, x squared over two, plus capital C, okay? Many students stop right here. But is capital C really free? No, it is fixed by the starting value! f of zero is one, and the natural log of one is zero, so capital C is zero. So f of x equals e to the power x squared over two. Beautiful, right?

2:49Type 2: both limits are constants

Now the second type. Same look, but the limits are now a and b, and both are constants. Okay? Here is the key question. When x changes, does this integral change? No, it does not! Inside there is only t, and t runs from a to b, two fixed numbers. So the whole integral is one number: we call it small c. Right? So f of x equals g of x plus c. Now we know the shape of f. Put it back into the integral, and you get one equation in c. Solve it, done. Got it?

3:21Differentiation under the integral sign

Now the third idea, and this one feels like magic. Suppose the integrand, capital F, depends on x and also on a second letter, a. We integrate in x, from p to q. On the left, d over d a means: differentiate this integral with respect to a. On the right, we simply take that derivative inside the integral. The curly d is a partial derivative: differentiate with respect to a, treating x as a constant. Okay? But think for a moment. What if the limits p or q also changed with a? Then this simple form is not enough, and we need the full Leibniz's rule from the previous concept. So this form is only for fixed limits, where p and q do not depend on a. Clear?

4:06Example 1: solve f(x) = x + integral 0 to 1 of t f(t) dt

Example one, from the course. Find f, if f of x equals x, plus the integral from zero to one of t times f of t, d t. Pause the video and try it first. Done? Okay, let's check. Which type is it? The limits are zero and one, both constants. So the integral is a number, call it c, and f of x equals x plus c. Okay? Now find c. Inside the integral we have t times f of t. And f of t is just t, so c is the integral of t times t, which is one third, right? Oh no, careful here! f of t is t plus c, not just t. The same c sits inside the integral too. This is the classic slip. With me? Open the bracket: t squared, plus c times t. The integral of t squared from zero to one is one third. And c comes out, times the integral of t, which is one half. So c equals one third plus c over two. Okay? Bring c over two to the left. c minus c over two is c over two, and that equals one third. So c is two thirds. Clear? So f of x equals x plus two thirds. Put it back into the question, and it checks out. That's it!

5:23Example 2: integral 0 to 1 of (x cubed - 1)/ln x is ln 4

Example two, our puzzle from the start. Evaluate the integral from zero to one of x cubed minus one, over the natural log of x, d x. The hint: use capital I of a, the same integral with x to the power a in place of x cubed. Pause the video and try it first. Done? Okay, let's check. Put a equal to three, and capital I of a becomes our integral. So we want I of three. The limits, zero and one, do not depend on a, so we may differentiate under the integral sign. Clear? Now differentiate the integrand with respect to a, keeping x fixed. What is the derivative of x to the power a, with respect to a? Yes, exactly! It is x to the power a, times the natural log of x. The minus one is a constant, so it gives zero. And look, the natural log of x on top cancels the one at the bottom. The ugly part is gone! Okay? So capital I prime of a is the integral from zero to one of x to the power a, d x. That is one over a plus one. Easy, right? Now integrate back with respect to a. The integral of one over a plus one is the natural log of a plus one. So capital I of a is the natural log of a plus one, and we are done? No, we are not done yet! We forgot the constant, plus capital C, okay? To find it, put a equal to zero: x to the power zero, minus one, is zero, so the integrand is zero and I of zero is zero. The natural log of one is also zero, so capital C is zero. With me? So our integral is I of three, the natural log of three plus one, that is, the natural log of four. Puzzle solved, with no antiderivative at all. Very good!

7:13Quick revision: x in a limit, constant limits, fixed limits

So, quick revision. If x sits in a limit, differentiate, and get the starting value by putting x equal to a. If both limits are constants, the integral is a number c: write f as g plus c, and solve for c. With fixed limits, differentiate inside with respect to a, integrate back, and fix the constant. Now go and try the practice problems on this page. All the best!