Notes - Matlan - Applications of Integrals
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Simple Area
Just remember signed area. The area under a curve, if the curve goes above the x axis is positive, while if it goes below the x axis (meaning, the area "under" is actually above) the area has a negative sign. To find the absolute unsigned area you need to find the area for multiple intervals.
Area Restricted by 2 Curves
L=(∫_a^b)([ƒ(x)-g(x)]*d(x))
Do note that signed area doesn't matter here due to the minus sign.
Ex. 1
Find the area restricted by:
The area is restricted by the intersection between the 2 curves. This means that we have to find the intersection first before we can solve, to know which curve to subtract from the other. We find that the intersections between the 2 curves are at (-2,-2) and (1,4). So, the interval is -2 to 1
In this case, the curve to subtract from is the second one, y=2*x+2, because for the duration of the entire interval the value remains higher than y=x2+3*x
L=(∫_-2^1)((2*x+2-(x2+3*x))*d(x))
=(∫_-2^1)((-x2-x+2)*d(x))
=(∣_-2^1)(-1/3*x3-1/2*x2+2*x)
=-1/3⋅(1-(-8))-1/2⋅(1-4)+2⋅(1-(-2))
=9/2
Ex. 2
Find the area restricted by:
At the interval 0≤x≤5/4*π
Notice how in the middle of the interval there is an intersection at x=π/4. In this case, the intersection shows that the "upper" curve swaps. So, we need to separate the solution into multiple definite integrals.
First, the we define the area area where cos(x) is the upper curve, then another area where sin(x) is the upper curve.
L=(∫_0^π/4)((cos(x)-sin(x))*d(x))+(∫_π/4^5*π/4)((sin(x)-cos(x))*d(x))
=(∣_0^π/4)(sin(x)+cos(x))+(∣_π/4^5*π/4)(-cos(x)-sin(x))
=(sin(π/4)-sin(0*π)+cos(π/4)-cos(0*π))+(-cos((5*π)/4)+cos(π/4)-sin((5*π)/4)+sin(π/4))
=(√(,2)/2-0+√(,2)/2-1)+(-(-√(,2)/2)+√(,2)/2-(-√(,2)/2)+√(,2)/2)
=-1+3√(,2)
Highlight: In this case I combined both areas from the start. You should probably not do that and compute both areas in 2 different variables in order to minimize mistakes.
Volume of Rotated Objects
V=(∫_a^b)(ƒ(x)2*d(x))
Imagine if the area was rotated around the x axis, V is the volume of that volume. That's if it's 1 curve.
If it's 2 curves, then the equation is different.
V=(∫_a^b)((ƒ(x)2-g(x)2)*d(x))
Ex. 1
Find the volume of the area rotated by the x axis restricted by:
Self explanatory. The interval is -1 to 2 and there are no intersections inside the interval. So, we find the volume directly, with y=x+3 being the upper curve.
V=(∫_-1^2)(((x+3)2-(x2+1)2)*d(x))
=(∫_-1^2)((x2+6*x+9-x4-2*x2-1)*d(x))
=(∫_-1^2)((-x4-x2+6*x+8)*d(x))
=(∣_-1^2)(-1/5*x5-1/3*x3+3*x2+8*x)
=-1/5⋅(2-(-1)5)-1/3⋅(2-(-1)3)+3⋅(2-(-1)2)+8⋅(2-(-1))
=-33/5-3+9+24
=117/5