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
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!
Latest Answers