I’m in CalcII now and for the life of me cannot wrap my head around integrals. Now we are using things like u-Substitution methods and solving the areas between 2 curves. I can understand how the equations work, but not why because I still cannot picture what an integral is or why it’s important.
In: 51
An integral takes a rate of change, and turns it into a graph (the opposite of a derivative). It also gives you the area under a given graph (because a graph is the rate of change of the area beneath it, no?).
This is incredibly useful when you’re dealing with something complex that’s changing in a particular way. In practical terms, an integral is just breaking down a complex problem into incredibly small pieces, solving those pieces individually, and bringing them back together; integrating the answers.
This video does a nice job at showing how to think of integrals: [https://youtu.be/WUvTyaaNkzM](https://youtu.be/WUvTyaaNkzM)
An integral is like the sum function (capital sigma), but rather than being discrete (eg. Considers function values/graph heights at x-positions 1, 2, 3 and 4) it’s continuous (eg. considers all function values between x-positions 1 and 4.)
Think about a line on a graph, it can be any, say y = x^2
We can get an alright guess to the area under this curve by placing rectangles at even intervals underneath it, with some arbitrary (constant) width and being just tall enough to match the curve), eg:
Place 1-unit wide rectangles along the x-axis, and give them height such that each just about meet your graph (y=x^2). It’s not gonna look great, but you can find the area by hand easily (it’s discrete – there’s a countable number of rectangles). Now make them each half as wide (0.5-units wide), and fill the gaps with more bars. takes more to compute, but it’s a better approximation.
Do this forever, and if you generalise the process, you get integration! It’s called integration by first principles. Gives you an exact expression for the area under a curve.
It’s the inverse operation of a derivative.
F'(x) = f(x)
∫f(x)dx = F(x)
It’s the amount of area under a given curve.
If we take a curve and break the area underneath it into several rectangles, the height of a given rectangle is f(x), and the width of the rectangle is Δx
The area of one of those rectangles would be f(x) * Δx
Now, we need to add all of those rectangles together.
If we split the area into n rectangles, and we integrate from a to b, so our rectangle heights are f(a), f(a+Δx), f(a+2Δx),… until f(b). The width of all the rectangles is Δx.
Δx = (b-a)/n
Now we just add them all together. Σ (i=0->n) f(a+i Δx)Δx
When we take the limit as n->infinity, then it becomes an integral
An integral creates a relationship between two semi-related dimensions. Say your x-axis is time and your y-axis is speed (or distance per unit time). When you take the integral of the data points presented (i.e. time * distance / time), you now have a new data point: distance.
If a device has an accelerometer, then you can track how far that device has moved without using other tech-expensive techniques like triangulation or GPS queries.
There’s countless gauges in smart-devices that rely on integrals to interpret data presented by gauges into something potentially more meaningful.
While it’s the opposite of derivative, undoing of derivative, it has another conceptual meaning: that is it’s the area under a curve and the x axis. Why is this important? Being able to calculate the area under curve gives a starting point for getting volume of weird shaped things such as a bottle. In higher calc classes(2, 3) the outward edge of the bottle can be modeled as a curve and that curve is rotated around an axes to get the volume. This where the power of calculus comes to play. A sphere, box, cylinder has volume formula, but what is the formula for any weirdly shaped stone? There is none. You have to use calculus.
Most abstractly? The integral is the anti-derivative. That’s literally it, algebraically, you just are doing the inverse of the derivative.
Geometrically, it can have a lot of interpretations, but the most common one for single-variable functions is that it’s the (signed) cumulative area between the x axis and the function curve.
Just curious, how did you learn how to do integrals without being taught the stepwise method of rectangles under a curve and making them skinnier and skinnier until they are infinitesimally-thin slices??
People keep saying “it’s the area under the curve,” but I have an example and a relation to algebra that may make it easier to understand what this really means.
Let’s say that you want to measure how much water is in a tank. Water is going into the tank at a rate of 10 liters per second. How much water is in the tank after 5 seconds? Easy, 50 liters, we can find this using y = 10*t; t=5; so y=50. This is just algebra.
The problem gets harder when water is going into the tank at a rate of 10*t liters per second. That is, the speed that water enters the tank depends on how much time has passed, and it’s increasing! At t=1 it is entering at a rate of 10L/s, at t=2 it’s 20L/s and so on. How much water is in the tank at t=5?
Well the key insight of integral calculus is that you can sum up all of the little differences over time to get the answer. So let’s say for the duration t = [0,1), it’s not flowing. t = [1, 2) it’s flowing at 10L/s. Sum these up and we get 0 + 10 + 20 + 30 + 40 = 100L. But wait! It wasn’t flowing at 0 the *entire* time in the range [0,1). It started at 0, but increased steadily to 10L/s by t=1. If I repeated this process for very very tiny differences in t, I would end up pretty close to the actual answer. This is where you will see something like [this picture](https://en.wikipedia.org/wiki/Riemann_sum#/media/File:Riemann_sum_(y=x^2).gif) in your textbook. This process is Riemann Sums.
Now this is where you can use the cool integral calculus techniques you are learning: the indefinite integral with respect to t of y = 10t is: 1/2*10*t^2 = 5*t^2 + c. After 5 seconds, we have 5*5*5 = 125 Liters in the tank.
Now what’s the “+ c” that always trips everyone up? That’s how much water you started with in the tank, set it to 0 if the tank was empty.
This is the “area under the curve,” when the curve is actually a line… notice that this is the exact formula for the area of a triangle: 1/2*w*h; where w = t, h = 10*t.
One of the most common applications you will see when you start applying calculus to physics is gravity, where acceleration vertically is -9.8m/(s^2). Your velocity is just all the little sums of gravity over time, so integrate with respect to t to get v=-9.8*t + c (c is just your initial velocity). Position is just all the little velocities over time added up, so integrate again to get position as y=-9.8/2*t^2 + c*t + y_0 (I renamed the new “c” to “y_0”, it’s your starting position).
An example, I jump off a diving board 100m in the air with an initial velocity of 1 m/s, how long until I hit the water?
y=0 (I hit the water), c=1 (I jumped up), y_0 = 100m (I’m on a diving board in the sky), solve for t: -9.8/2*t^2 + t + 100 = 0, use quadratic formula (a=-4.9, b=1, c=100): t is either −4.41665
or 4.62073. I can’t go back in time, so t=4.62073
What if I didn’t jump? Quadratic formula again, this time with b=0, 4.51754. Jumping gave me about 0.11s of extra hang time. Regardless, I’m out of shape, so I’m still dead when I hit the water.
Hope this helps to make things a little more clear, and sorry for the formatting!
Many have posted the “Area under/between two wave forms” explanation, and yes. But — how does that work? This is how it was explained to me.
1. You have a wave (or two) and you can measure the Y distance at any X value.
2. Make a list of those values over a range. Consider those to be a bar graph with the bars having width of your X interval.
3. You can calculate the area of each bar (X * Y) and add them all up for a summary value. This would be (almost) the exact area between the two waves for any X to X+N. Except:
4. Unless both wave lines are parallel, there’s a crossing component at the end of each bar, making them trapezoids.
5. You could try again, calculating the areas of all those little upper and lower triangles, but it would still be a bit off because the wave lines are curved!
6. So — The Integral is the complete summation of all the vertical bars (infinitely narrow) including all the adjustments for the area of the little triangles (also infinitely small) at the ends.
Hope that helps!
An integral is like a math genie that measures curves and tells you how much water’s in your tank! 🧞♂️💧
Latest Answers