Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Integrals DYDX equals some function F of X. Then Y of X equals the integral of F of XDX plus C. This is referred to as the general solution. If G prime of X is equivalent or identical equality for all F of X, that three symbol = means identical equality for all X. Then we could think of Y of X is equaling the function G of X + C because by the fundamental theorem of calculus, an integral and a derivative undo each other and give us back the original function. If we know a given condition, then we can find a particular solution by solving for C So we're going to emphasize general solutions and particular solutions. General solution means we don't know what the C is. Particular solution means that we have some given condition where we can actually evaluate C If X equal F of T is used to describe a particles position along a straight line, then V of t = F prime of T or we could think of that notation DXDT is the velocity of the particle. Acceleration is a of T equaling V prime of T which equals X double prime of T, or a equal DVDT which equals the second derivative of X. In terms of time. Newton's law of motion states that force equals mass times acceleration. So we could think of that as force equaling M times the second derivative of X, or F / m equaling the second derivative of X. Constant acceleration. Force equals mass times acceleration, so acceleration equals force divided by mass, where a is a constant. So we know that A was just DVDT. So if we evaluate the integral, we'd get V equaling the integral ADT. But a is a constant, so that's going to equal AT plus some C, Let's call it C sub one. If we know that V naughty equals V at time zero, so the initial velocity is the velocity at time 0, then V naughty is going to equal a time 0 + C one. Or we can see that C1 is just the initial velocity. So now we've developed an equation of velocity equal acceleration times time plus the initial velocity. Building on that, we're going to have V equaling the AT plus V naughty. But the velocity was just the first derivative of the position. So if we evaluate the integral of both of those, we get the integral of DX DT equaling the integral of AT plus V naughty DT. So X = 1/2 AT squared. Remember A is still a constant plus V naughty t + C sub 2V naughty is also a constant. So if we know our initial position iex naughty equaling X at times 0, we can figure out our C2. So we have X naughty equaling 1/2 a times Z ^2 * 0 ^2 + v naughty times 0 + C two or X naughty is just equaling to C2. Thus X = 1/2 AT squared plus V naughty t + X sub naughty. Also we want to recall that speed is just the absolute value of velocity. Weight of a body is the force exerted on the body by gravity. So if we let A equal G&F equal W thus the F equal MA equation becomes W equal MG. If we think of ground level being y = 0 and upward as the positive direction, then gravity is decreasing its height and velocity. Thus, our DV DT is going to equal a negative G and thus our V is going to equal negative GT plus V naughty our Y of T. We usually use Y for heights instead of XS, so our Y of T is going to equal -1 half GT squared plus V naughty t + y naughty. These are basically the same equations that we developed a moment ago. But now because gravity is affecting the height and the velocity, our a is going to be a negative. Let's do some examples. If we have DY over DX equaling X sqrt X ^2 + 9, when Y of -4 = 0, we're going to take the integral of each side. So we're going to get Y equal the integral of X sqrt X ^2 + 9 DX. Now you may not need to use substitution, but if you don't see how to evaluate that integral, we can do AU sub where we have U equaling X ^2 + 9. So the DU would equal 2 XDX or 1/2 DU is XDX. We can see here is the X and here is the DX. So I'm going to take out the XDX and I'm going to replace it with the 1/2 DU. That still leaves me sqrt X ^2 + 9. But X ^2 + 9 we decided was AU. So that turns into the square root of U. So now we're just going to use the power rule. We're going to add 1 to the exponent and divide by the new exponent. So we get 1/3 U to the three halves plus C If we take out EU and substitute back what it equaled, we get 1/3 the quantity X ^2 + 9 to the three halves plus C So this is our general statement, but we had an initial value up here of Y of -4 equaling 0. So we're actually going to stick in -4, and we're going to figure out what the C is. So sticking in -4 for our X and 0 for our Y solving we can see is negative. We can see that C is -125 thirds, so Y equal 1/3 the quantity X ^2 + 9 to the three halves -125 thirds. We're going to recall that the integral of 1 / a ^2 + U ^2 DU is just really 1 / a tangent inverse U / a + C. If we don't recall that, we can make ourselves a right triangle and label our sides appropriately. Because it's a ^2 + U ^2, we're going to have EU being on the opposite side and the a is just a constant, so we're going to have it be on the adjacent so that we could have tangent Theta being U / a. If I cross multiply, I get a tangent Theta equaling U or a secant squared Theta D Theta equaling DU. So now when we evaluate this integral, we'd have 1 / a ^2 plus instead of U we're going to stick in a tangent Theta and we're going to square that times. Instead of DU we're going to stick in a secant squared Theta D Theta. When we do some simplification, we can see that we get a secant squared Theta over a ^2 1 plus tangent squared Theta D Theta. One plus tangent squared Theta is a Pythagorean identity and it's really just secant squared Theta. So the secant squared thetas are going to cancel and A / a ^2 is going to just leave US1 over AD Theta. Remember A is a constant, so the integral of one over AD Theta is going to turn into 1 / a Theta plus C And if we look up here, Theta is really just tangent inverse of U / a. So we get 1 / a times Theta plus C turning into 1 / a tangent inverse U / a + C. Doing similar types of triangles, we can see that the integral of DU over sqrt a ^2 -, U ^2 is sine inverse of U / a + C and also the integral of DU over U square root U ^2 -, a ^2 equal 1 / a secant inverse the absolute value U / a + C. We can develop these last two and similar methods by choosing a triangle and labeling the sides appropriately. Let's do another example. If we're given a of T equaling 2T plus one, V not equaling -7, and X not equaling 4, we get V equal the integral of 2T plus one DT, which would just be t ^2 + t + C one. That's just using the power rule. Add 1 to the exponent divided by the new exponent. Putting in the original value, we're going to have -7 equaling 0 ^2 + 0 plus C1. Remember that V naughty of -7 really is saying that the velocity at zero is -7. So now we can see that the C1 turns into -7. Now if we wanted to find the X, we're going to have X equaling the integral of that velocity equation. So we get one third t ^3 + 1 half t ^2 -, 7 T plus C2. Once again putting in our initial value, we know that X not equal 4 or X at time 0 is location 4. So we put 4IN for our X0 is in for our T We solve for C sub two and we get that's being 4. So our final equation for position as X equal 1 third t ^3 + 1 half t ^2 -, 7 T +4 forgiven a graph of velocity, and we want to find the graph of the resulting position function. When we look at this, we can see that this is going to be a straight line, so our velocity is just going to be T We've gone up and over, up and over an equivalent amount. Remember that's .00. We can also see that the velocity from 5 to 10 is going to be 5. So if I know my velocity equaling T from zero to five and five from 5:00 to 10:00, I can find my X by taking the integral. So we get one half t ^2 + C one and 5T plus some C2. Well, we know on our .00. Thus we know that 0 = 1 half 0 ^2 + C one. So we can show that C1 is 0. We also know that the .55 is got to be in that first integral and hence when we're going to figure out what the X value is when the time was 5 S X = 1 half 5 ^2 or 25 halves. So now because the .55 is on both of those equations, we're going to actually substitute the 25 halves in for our X location to find our C2SO25 halves equals 5 * 5 + C two or we can see that C2 is -25 halves. So now coming up with equations, we get X equal 1 half t ^2 for zero to five and five t -, 25 halves from 5:00 to 10:00. If we go ahead and graph that, that top from zero to 5 is a parabola and from 5:00 to 10:00 is a straight line. I found the 75 halves literally by sticking 10 in here. So 50 -, 25 halves. Thank you and have a wonderful day.