Overview

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Take the integral from zero to infinity of d x over, one plus x squared, times one plus x to the power two thousand twenty six. Partial fractions, with a power that big? That looks impossible, right? By the end of this video, you will do it in four short lines. Here the questions are mixed, so nobody tells you which tool to use. So, let's begin!

0:29How this chapter is built

Where are the real exam questions? The JEE Main questions from twenty twenty one to twenty twenty six sit at the end of Chapters one to six; the short dash means to. Okay? This chapter has original problems in three mixed sets: nine solved plus two practice problems each. Mixed means any chapter can come next, so you spot the type. Got it?

0:52Indefinite integrals: never forget +C

First family: indefinite integrals. The long S is the integral sign, and d x says we integrate in x. The integral of f of x is capital F of x plus capital C, where F prime, with the little dash, equals f. Never drop the constant. Okay? Your tools from Chapters one and two: substitution, parts, and partial fractions, which need a proper fraction; otherwise, divide first. Right? One pattern to spot: the integral of e to the power x, times f plus f prime, is e to the power x times f, plus capital C, okay? From Set one: e to the power x, times tan x plus sec squared x. Which f fits? Tan x, exactly! Its derivative is sec squared x. So the answer is e to the power x times tan x, plus capital C, with no integration by parts needed. Clear?

1:49Definite integrals: look for a property first

Second family: definite integrals, with limits a and b on the sign. The integral from a to b is capital F of b minus capital F of a, when f is continuous on the closed interval from a to b; the square brackets include both ends. Okay? But before integrating, look for a property. The King property: the integral from a to b of f of x equals the integral of f of a plus b minus x, over the same limits. Right? For symmetric limits, minus a to a: an even f gives twice the integral from zero to a, and an odd f gives zero. Even means f of minus x equals f of x; odd means it equals minus f of x. With me?

2:32Leibniz's rule: limits that move

Third family: Leibniz's rule, for limits that move. d over d x, the derivative, of the integral from g of x to h of x of f of t, d t, is f of h of x times h prime of x, minus f of g of x times g prime of x. It needs f continuous, and g and h differentiable. Clear? From Set one: d over d x of the integral from zero to x cubed of e to the power t squared, d t. What do you get? Put x cubed in place of t: e to the power, x to the power six. Then multiply by the derivative of x cubed, three x squared; the dot means times. The constant lower limit adds nothing. Forgetting the three x squared is the usual slip, okay?

3:19Limits of sums become integrals

Fourth family: limits of sums. In the figure, under the curve y equals one over one plus x squared, each rectangle has width one over n and height f of r over n. The capital Greek sigma adds them, r from one to n. Lim means limit: n tends to infinity, growing without end; the sideways eight is infinity. The rectangles then fill the area, the integral from zero to one. With me? The arrows mean becomes: r over n becomes x, one over n becomes d x, and the range of r over n gives the limits. Right? From Set one: the limit of root one plus root two, up to root n, over n to the power three over two; the tick is the root sign, and the dots mean and so on. Can you write it in this form? Write n to the power three over two as n times root n. Each root r over root n is root of r over n, with one over n outside. So we get the integral from zero to one of root x: two thirds. Easy, right?

4:23Area is never negative

Last family: areas. In the figure, the integral from zero to two pi of sine x is zero. So the area between the curve and the x axis is zero too, right? No, no, careful here! This trap catches many students. The integral counts the hump below the axis as negative: plus two above, minus two below, and they cancel. But area is never negative. So take mod sine x, with the vertical bars, and split where the curve crosses the axis, at pi. Two plus two: the area is four, just as the figure says. Okay? Between two curves: the integral of y upper minus y lower, d x; the small words name the curve, and a and b come from the intersection points. Always sketch first. Clear?

5:11Example 1: ln tan x and sin 2x ln tan x via King property

Example one, from Set one, marked easy. The integral from zero to pi over two of natural log of tan x; the letters l n mean natural log. And the same integral of sine two x times natural log of tan x. Pause the video, and try it first. Done? Okay, let's check. Many students try by parts here, and get stuck. So which is it: by parts, or a property? Yes, a property! The limits are zero to pi over two, so use the King property: the arrow means replace x by pi over two minus x. Tan of pi over two minus x is cot x, one over tan x. With me? And natural log of one over tan x is minus natural log of tan x. So the first integral, capital I, equals minus itself; the double arrow means implies. So two I is zero, and capital I is zero. Okay? For the second integral, capital J, sine two x becomes sine of pi minus two x, which is sine two x again. Only the log changes sign, so capital J equals minus itself: zero too. Two integrals, and almost no integration!

6:24Example 2: dx/((1+x2)(1+x2026)), 0 to infinity, x to 1/x

Now, remember our puzzle? Let's finish it. From Set three, marked hard: the integral from zero to infinity of d x over, one plus x squared, times one plus x to the power two thousand twenty six. Pause the video, and try it first. Done? Okay, let's check. Partial fractions are hopeless here. Look at the limits, zero to infinity. Which substitution swaps zero and infinity? x goes to one over x, exactly! Then d x becomes minus d x over x squared, and the limits change: zero becomes infinity, infinity becomes zero. The minus sign flips them back. Okay? Multiply top and bottom by x squared, times x to the power two thousand twenty six. The bottom becomes the same two brackets, and the top is x to the power two thousand twenty six. So capital I also equals this. With me? Add the two forms: the tops make one plus x to the power two thousand twenty six, which cancels the bracket below. Two I is the integral from zero to infinity of d x over one plus x squared: tan inverse x, the small minus one meaning inverse, in square brackets from zero to infinity. Pi over two. Right? So capital I is pi over four. Four short lines, and the huge power simply cancelled. Puzzle solved!

7:54Quick revision: plus C, property first, Leibniz, sums, area

So, quick revision. Indefinite: plus capital C every time. Definite: look for a property first. Moving limits: Leibniz's rule. Sums: one over n times the sum becomes an integral from zero to one. Area: sketch, split where the curve crosses the axis, never negative. Now open Set one and start solving. All the best!