## Introduction: Manual Derivatives and Integrals

This procedure will guide you through some of the core techniques of calculus: derivatives and integrals. These are essential for almost any kind of math or science career. You'll need a good grounding in algebra to understand the sample equations, and writing materials to check your work. Going through these steps could range from minutes to hours depending on how new you are to calculus, but even if you're an experienced mathematician, you should pace yourself and try to contemplate the fundamental theory of what happens in each step. Although people usually study it in high school at their earliest, calculus was discovered over 300 years ago, so any practiced math student can learn it.

## Step 1: Find a Derivative

1. Find or make a function of one variable. The function will give an output that we can use as the second variable in a two-dimensional graph, which in most cases means f(x)=y.

2. Use Newton's difference quotient. For a small segment of a function graph, with width h, slope is found with m=[f(x+h)-f(x)]/h, taking the limit of this equation at h approaches 0.

3. Work out the math. In the example polynomial, we replace all instances of (x) with (x+h) to get m=[2(x+h)^3 + 4(x+h)^2 +3(x+h) + 5 - 2x^3 - 4x^2 - 3x - 5]/h. This works out to m=6x^2 + 6xh + 8x + 2h^2 + 4h + 3, as h approaches 0. Thus, the slope of the expression 2x^3 + 4x^2 + 3x + 5 is given at any x value by 6x^2 +8x + 3. Most people studying derivatives for the first time will find it helpful to fully work out the algebra. If you still struggle, step-by-step guidance can be found with a tutor or www.wolframalpha.com.

4. Generalize the results to find derivation rules. As you may notice with the previous example, each term's exponent decreases by one in the derived polynomial; there are derived rules displayed above for a few general cases that have been studied numerous times.

## Step 2: Riemann Sums

1. Find a function whose (x,y) coordinates you can calculate over an interval. Common intervals usually start at or center around x=0.

2. Divide the area under the curve into reasonably small rectangular segments of uniform width. For example, 0.1x units apart.

3. Approximate the height of each segment with the y value of the function halfway between its high and low x values. In some circumstances, you may specifically want an approximation that's strictly too high or too low, prompting you to use a different y value from the curve segment.

4. Find the area of each segment by multiplying height by width. You will need to record these individual areas for the next step.

5. Add the areas to find total area under the curve for the interval. This is a large amount of mathematical grunt work, but it is reliable and very nearly approximates the total area, especially with more narrow segments. This is how most calculators perform integrals.

## Step 3: Definite Integration

1. Find a simple polynomial, such as ones in the previous steps.

2. Find a derivation rule that could produce the function. As there are numerous types of functions and infinite examples for each type, as well as combinations of types, along with some functions that are impossible to integrate, you can't exhaustively explore the options for integration; in this case, the answer should be straightforward.

3. Reverse the process of a derivation. For a polynomial term, increase the exponent by one and divide the coefficient by the new exponent.

4. Account for a constant of integration. When you find the derivative of a polynomial that gives y as a function of x, the final term, which is in the power of x^0, is lost. A given polynomial derivative can come from an infinite number of parent polynomials. However, all coordinates are shifted by the same constant, so the next step still works.

5. Find the difference between the ends of the integrated function. Because the original function, this new function's derivative, gives the series of slops between these ends, this difference equals the area under the original function's curve. This relationship is largely unintuitive, so try working out as many examples as needed to observe it.

## Step 4: Physical Applications: Forward Acceleration and Deceleration

1. Imagine a car that starts at rest, accelerates for 5 seconds at 2ft/s^2, then runs steadily at 10 ft/s for 5 seconds, then slows down at a rate of 1ft/s^2 until it stops.

2. Try to find the distance from the starting point mentally. This should be mildly tricky, since the rate of motion changes over time. The only easy parts to solve mentally are the distance traveled at a constant velocity and the time it takes to stop (10-1t=0, so t=10). You may check your answer below, then try the mathematical method.

3. Break the story problem into a piecewise function. The velocity and time can be represented as v=2t for 0 to 5 t, v=10 for 5 to 10 t, and v=20-t for 10 to 20 t.

4. Integrate the pieces separately with respect to the variable t.

• v=2t: The exponent must be one degree higher and the coefficient must be divided by the new exponent. Since the exponent is implied to be one, we get x=t^2 over the t interval from 0 to 5. This gives x=(5)^2-(0)^2=25.
• v=10: Velocity is constant, but we can use this step to check that our reasoning is sound. The exponent rule gives x=10t, given t from 5 to 10, which is x=10(10)-10(5)=50.
• v=20-t: Integrate each term separately; x=20t - 0.5t^2 from 10 to 20, which is x=20(20) - 0.5(20)^2 - [20(10) - 0.5(10)^2]=50.

5. Compare the sum of the integrals to the graph. Note that the vertical scale is 5 feet per block.

## Step 5: Physical Application: Malfunctioning Rocket

1. Try integrating a higher order polynomial. If you were very familiar with story problems like the accelerating car, you might have known that you can graph ft/s vertically and seconds horizontally to make a graph that's very easy to find the area under. However, a third- or fourth-degree polynomial will not be so visually simple even when you use this kind of method.

2. Imagine a rocket that launches from the ground and has a brief fuel jam, so that its velocity in ft/s at a given time is given by v=6t^2 - 16t + 8 for the first minute of its flight, after which it hits a bird and expels all fuel horizontally, while the only force acting on it is gravity; find the time at which it lands.

3. Assess the problem. Since neither velocity nor acceleration is constant, the only way to solve the problem is to integrate for the position function, find the change in position for the first minute, and find how long an object in free fall from that height takes to land.

4. Execute your strategy. Integrating each term of the polynomial separately, we get h=2t^3 - 8t^2 + 8x + C for the height in feet. We then plug in the endpoints, t=60 and t=0 to get [2(60)^3 - 8(60)^2 +8(60)] - = 432000-28800+480=403680. After this point, the rocket falls with an acceleration of 32 ft/s^2 to the Earth. Since acceleration is constant, we can easily integrate a=32 into v=32t and then h=16t^2, which we equate to the maximum height to find how long it spends falling. 16t^2=403680 --> t=158.8. Adding 60 seconds to this, the rocket hits the ground 218.8 seconds after it launches.

## Step 6: Conclusion

You have now learned some of the basic tools for higher mathematics and science. From here, your improved understanding of calculus should allow you to perform well in more advanced studies such as field physics, thermodynamics, and quantum mechanics. If you have any problems, remember there's nothing wrong with seeking out a tutor or using Wolfram Alpha to walk you through the steps of a complicated problem.