# Integral Calculus

## Contents

Integral Calculus Basics:

Problem Solving using Integral Calculus:

## Fundamental Theorem of Calculus

The fundamental theorem of calculus explains how to evaluate definite integrals of functions that have indefinite integrals. It bridges the concept of an antiderivative with the area problem.

When you figure out definite integrals (which you can think of as a limit of Riemann sums), you’re probably aware of the fact that the definite integral is just the area under the curve between two points (lower and upper limits). You are finding an antiderivative at the upper and lower limits of integration and taking the difference. The Fundamental Theorem of Calculus justifies this procedure.

The technical formula is:

and

The first part of the fundamental theorem is stating that when evaluating indefinite integrals between two points a and b, simply subtract the value of the integral at a from the value of the integral at b. The second part of the theorem gives an indefinite integral of a function. It looks complicated, but all it’s really telling you is how to find the area between two points on a graph.

## Fundamental Theorem of Calculus Example

Sample problem: Evaluate the following integral using the fundamental theorem of calculus:

Step 1: Evaluate the integral. This particular integral is evaluated using the integral rule for power functions:
∫x2 dx = x33

Step 2: Find the value of the integral at b, which is the value at the top of the integral sign in the problem. In this example, the value of “b” is 1, so:
x33 = 133 = 13.

Step 3: Find the value of the integral at a, which is the value at the bottom of the integral sign in the problem. In this example, the value of “a” is 0, so:
x33 = 033 = 0

Step 4: Subtract a (Step 3) from b (Step 2):
13-0 = 13

That’s it!

## Integral Calculus: Indefinite Integral Rules

Integration is the reverse process of differentiation. Integrals may or may not exist, but when they do, there are some general rules you can follow to simplify the integration procedure.

## Sum, Multiplication and Difference Indefinite Integral Rules

### The Sum Rule

The integral of the sum of two functions is the sum of their separate integrals:

∫[f(x) + g(x)]dx = ∫f(x)dx + ∫g(x)dx
∫[f(x) − g(x)]dx = ∫f(x)dx − ∫g(x)dx

### The Multiplication Rule

Any constant factor can be moved outside of the integration symbol:

∫axndx = a∫xndx for any constant ‘a’.

Similarly,
∫[af(u) + bg(u)]du = a = ∫ f(u) du + b ∫g(u) du

### Integral Power Rule

∫xndx = xn + 1n + 1 + c, Where n ≠ 0

### Integration by parts

∫u dv = uv – ∫v du

### Basic Integrals

∫m dx = mx + c, for any number m.
∫xn dx = 1n + 1xx + 1+ c, if n ≠ –1.
1xdx = ln |x| + c, for x ≠ 0.
∫sin x dx = −cos x + c
∫cos x dx = sin x + c
∫ex dx = ex + c

### Linear Substitution

If F'(x) = f(x) then for any m ≠ 0,
∫f(mx + b)dx = 1mF(mx + b) + c

### Indefinite Integrals for Trigonometric Identities

∫ cos x dx = sin x + C
∫ sin x = -cos x + C
∫ sec2 x dx
∫ csc2 x dx = -cot x + C
∫ sec x tan x dx
∫ csc x cost x dx

## Integral Calculus: Finding definite integrals

Definite integrals in integral calculus produce a result (a number that represents the area) as opposed to indefinite integrals, which are represented by formulas. While Riemann sums can give you an exact area if you use enough intervals, definite integrals give you the exact answer — and in a fraction of the time it would take you to calculate the area using Riemann sums (you can think of a definite integral as being an infinite amount of intervals in a Riemann sum).The definite integral is also known as a Riemann integral (because you would get the same result by using Reimann sums).

### Formal definition for the definite integral:

Let f be a function which is continuous on the close interval [a,b]. The definite integral of f from a to b is the limit:

Where:

is a Riemann sum of f on [a,b].
The formal definition of a definite integral looks pretty scary — but all you need to do is to calculate the area between the function and the x-axis. You’ll need a solid understanding of the general rules of integration for indefinite integrals in order to calculate a definite integral.

## Example

Sample problem: Calculate the area between x=0 and x=1 for f(x)=x2.
Step 1: Set up integral notation, placing the smaller number at the bottom and the larger number at the top:

Step 2: Find the integral, using the usual rules of integration:

Step 3: Substitute the top number for x and then solve:

Step 4: Add a subtraction sign and then substitute the bottom number for x, solving the integral:

That’s it!

Tip: There in another kind of definite integral, called a contour integral. Contour integrals involve those definite integrals that are taken on the complex plane (as opposed to the real line). However, you’re unlikely to encounter contour integrals in elementary calculus classes.

## Integral Calculus: Difference between proper and improper integrals

Improper integrals are integrals that can’t be evaluated as they first appear, while you can easily integrate a proper integral as is.

Integrals can be solved using a variety using a variety of techniques, including the power rule, integration by parts and substitution techniques like trigonometric substitution. When you integrate, you are technically evaluating using rectangles with an equal base length (see how to use Riemann sums to evaluate the area under a curve). You’re taking a known length (for example from x = 0 to x = 20) and dividing that interval into a certain amount of tiny rectangles with a known base length (even if it’s an insignificantly tiny length). However, if your interval is infinite (because of infinity being one if the interval ends or because of a discontinuity in the interval) then you start to run into problems. If you don’t know the length of the interval, then you can’t divide the interval into n equal pieces. If you can’t divide the interval, you have an improper integral.

What could cause you to not know the interval length? One reason is infinity as a limit of integration. Another common reason is that you have a discontinuity (a hole in the graph).

## Proper and Improper Integrals: Sample Problem

Sample problem: Figure out if the following integrals are proper or improper:
1.

2.

3.

4.

5.

Step 1: Look for infinity as one of the limits of integration. If infinity is one of the limits of integration then the integral can’t be evaluated as written. Sample problems #1 and #3 have infinity (or negative infinity) as one or both limits of integration.

Step 2: Look for discontinuities, either at the limits of integration or somewhere in between. This step will require you to use your algebra skills to figure out if there’s a discontinuity or not. Alternatively, graph the interval and look for asymptotes. Sample problem #4 has a discontinuity at x=9 (at this point, the denominator would be zero, which is undefined) and sample problem #5 has a vertical asymptote at x=2. Therefore, they are both improper integrals.

That’s it! The remaining integral (sample problem #2) is a proper integral because it is continuous over the entire interval.

Tip: In order to evaluate improper integrals, you first have to convert them to proper integrals.

## Integral Calculus: Improper Integrals

Improper integrals are integrals you can’t immediately solve because one of the limits of integration (or both limits) are at infinity or there is a vertical asymptote in the interval. The reason you can’t solve these integrals without first turning them into a proper integral is that in order to integrate, you need to know the interval length. And if your interval length is infinity, there’s no way to determine that interval. The workaround is to turn the improper integral into a proper one and then integrate by turning the integral into a limit problem. Limits for improper integrals do not always exist; An improper integral is said to converge if the limit exists and diverge if it doesn’t.

## Solve Improper Integrals in Integral Calculus: Steps

Sample problem #1:Integrate the following:

Step 1: Replace the infinity symbol with a finite number. For this sample problem, use “b” to replace the upper infinity symbol.

Step 2: Integrate the function using the usual rules of integration. The integral of 1x2 is -1x, so:

Step 3: Evaluate the definite integral:

As b approaches infinity, -1/b tends towards zero. That should be clear by looking at a table:

 b -1/b 1 -1/1 10 -1/10 100 -1/100 1000 -1/1000

Therefore, the limit -1b + 0 becomes 0 + 1 = 1.

That’s it!

How to Solve Improper Integrals Sample problem #2: Integrate the following:

Step 1: Replace the infinity symbol with a finite number. For this sample problem, use “b” to replace the upper infinity symbol.

Step 2: Integrate the function using the usual rules of integration. The integral of 1/x is ln|x|, so:

Step 3: Solve the Definite Integral:

As b tends towards infinity, ln|b| also tends towards infinity. This should be clear by making a table:

 b ln|b| 1 0 10 ≈2.3 100 ≈4.6 1,000 ≈6.9 10,000 ≈9.21 100,000,000 ≈18.42

Therefore, the integral diverges (it does not exist).

That’s it!

## Integral Calculus: Integration by Parts

Integration by parts is a technique used in integral calculus when you have integrals that are a product of two functions. For example, you would use integration by parts for ∫x*ln(x) or ∫xe5x. The formula for integration by parts is:

It looks complicated, but all it requires you to do is to split your equation into two parts, then differentiate them and multiply.

Sample Problem #1:Integrate x sin 2x using integration by parts.

Step 1: Identify u and v from the problem. With few exceptions, u will be the first part of the equation and v will be the second part of the equation:
u = x
v = sin 2x

Step 2: Multiply u with the integral of v:
= (x)* (-12 cos 2x)
Set this aside for a moment.

Step 3: Multiply the integral of v by the integral of u:
= (-12 cos 2x)(1)

Step 4: Combine your answers from Step 2 and 3, placing “-∫” between the two parts and “dx” at the end:
= (x)* (-12 cos 2x) -∫ (-12 cos 2x)(1) dx

Step 5: Simplify, bringing constants out in front of the integral:
= –x2 cos 2x + 12∫(cos 2x) dx

Step 6: Perform the last integration. In this case, integrate (cos 2x):
= –x2 cos 2x + 12(12sin 2x) dx

Step 7: Add “C” and then simplify:
= –x2 cos 2x + 14(sin 2x) dx

That’s it!

## Integral Calculus: Indefinite Integrals of power functions

An antiderivative, F(x) is the compliment of a function f(x); You can differentiate F(x) to get f(x), or antidifferentiate (integrate) f(x) to get F(x). Graphically, antiderivatives are the set of all vertical transformations of the antiderivative. For example, the antiderivative of 2x is x2 + C, where C is a constant. The derivative of a constant is zero, so C can be any constant, positive or negative. For example, four antiderivatives of 2x are x2 + 1, x2 -1, x2 + 2 or x2 – 2.

The following general rule is for integrating power functions of the form f(x)=xn[n ≠- 1):

It’s actually easier than it looks — all the formula is saying is to add one to the power, divide by that power, and then add a “C” for constant.

## Antiderivatives of Power Functions: Sample Problems

Sample Problem #1: Find the antiderivative (indefinite integral) for 20x3

Step 1: Increase the power by 1:
20x3=20x4

Step 2: Divide by the new power you calculated in Step 1:
20x4=204 x4 = 5x4

5x4 + C

Sample Problem #2: Find the antiderivative (indefinite integral) for 3x8

Step 1: Increase the power by 1:
3x8=3x9

Step 2: Divide by the new power you calculated in Step 1:
3&#82609x9 = 1&#82603x9

1&#82603x9 + C

Sample Problem #3: Find the antiderivative (indefinite integral) for x4+6

Step 1: Increase the power by 1 for x (note that you add x0 to a constant on its own — in this case, 6 becomes 6x0).
x4 + 6x0=x5 + 6x1

Step 2: Divide by the new powers you calculated in Step 1:
x5&#82605+6x1&#82601 =1&#82605x5+6x1

1&#82605x5+6x1+C

That’s it!

Tip: When adding 1 to your power, remember that x becomes x1, and a constant becomes that constant plus x0. For example, 6x10 + 17x + 9 becomes 6x10 + 17x1 + 9x0 before you add the power and 6x11 + 17x2 + 9x1 after you add the one.

## How to find the area between two curves in integral calculus

Finding the area between two curves in integral calculus is a simple task if you are familiar with the rules of integration (see indefinite integral rules above). The easiest way to solve this problem is to find the area under each curve by integration and then subtract one area from the other to find the difference between them. You may be presented with two main problem types. The first is when the limits of integration are given, and the second is where the limits of integration are not given.

## Area Between Two Curves: Limits of Integration Given

Sample problem 1: Find the area between the curves y = x and y = x2 between x = 0 and x = 1.

Step 1: Find the definite integral for each equation over the range x = 0 and x = 1, using the usual integration rules to integrate each term. (see: calculating definite integrals).

Step 2: Subtract the difference between the areas under the curves. You’ll need to visualize the curves (sketch or graph the curves if you need to); you’ll want to subtract the bottom curve from the top one. The curve on top here is f(x)=x, so:
1213 = 16.

## Limits of Integration NOT Given

Sample problem: Find the area between the curves y = x and y = x2.

Step 1: Graph the equations. In most cases, the limits of integration will be clear, especially if you’re using a TI-calculator with an Intersection feature (just find the intersections of the two graphs). If you can find the intersection by graphing, skip to Step 3.

Step 2: Find the common solutions of these two equations if you cannot find the intersection by graphing (treat them as simultaneous equations).
Substituting y = x for x in y = x2 gives an equation y = y2, which has only two solutions, 0 and 1.
Putting the values back into y = x to give the corresponding values of x: x = 0 when y = 0, and x = 1 when y = 1. The two points of intersection are (0,0) and (1,1).

Step 3: Complete the steps in Sample Problem 1 (limits of integration given) to complete the calculation.