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By: Zack Broadhead.  When you take the derivative of a number with no variable it is just ‘0’.  Y=5x+4, (dy/dx)=5+0  When you antidifferentiate an.

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Presentation on theme: "By: Zack Broadhead.  When you take the derivative of a number with no variable it is just ‘0’.  Y=5x+4, (dy/dx)=5+0  When you antidifferentiate an."— Presentation transcript:

1 By: Zack Broadhead

2  When you take the derivative of a number with no variable it is just ‘0’.  Y=5x+4, (dy/dx)=5+0  When you antidifferentiate an equation the result must have a “+ C” on the end because you don’t know what the constant number was.

3  Integrate equation.  Tack “+C” onto the end.  Plug in initial values.  Solve for “C”.  Plug “C” into integrated equation.

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5  ∫ (dy)=∫x dx  Y=x 2 /2+C  Y=3 and x=2 so:  (3)=(2) 2 /2+C  C=1  Y=x 2 +1

6  ∫(dy)=∫Cos(x) dx  Y=Sin(x)+C  Y=1 and x=п/2 so:  1=Sin(п/2)+C  C=0  Y=Sin(x)+0

7  ∫(dy)=∫(x 2 +x) dx  Y=(x 3 /3)+(x 2 /2)+C  Y=1, and x=1 so:  1=1/3+1/2+C  1=5/6+C  C=1/6  Y=(x 3 /3)+(x 2 /2)+1/6

8  ∫(dy)=∫(1/x) dx  Y=ln(x)+C  Y=2 and x=1, so:  2=ln(1)+C  C=2  Y=ln(x)+2

9  ∫(dy)=∫(Sin(2x)) dx  Y=-(Cos(2x))/2+C  Y=1/2 and x=0, so:  ½=-(Cos(2(0)))/2+C  C=1  Y=-Cos(2x)/2+1

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11  A. y=x 3 -2  May have accidentally subtracted “C”.  B. y=6x-2  May have derived instead of integrated.  C. y=x 3 +2  Correct answer!  D. y=x  Plugged wrong points in.

12  A. y=√((4/5)x 5 –(123/5))  Used wrong points.  B. y=√((4/5)x 5 +(16/5))  Correct!  C. y 2 /2=(2/5)x 5 +(8/5)  Not simplified.  D. y=8x 3 -6  Derived instead of integrated.

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14  Answer: Y=x 2 -x+5  ∫(dy)=∫(2x-1) dx  Y=x 2 -x+C  Y=11 and x=3, so:  11=(3) 2 –(3)+C  C=5  Y=x 2 –x+5

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16 By: Zack Broadhead


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