https://padlet.com/kmbeal/upqoa6tah5e7 Post a screenshot of a function and discuss the steps to solving a definite integral over three different intervals. Discuss and critique on other classmate's posts. en-us 2019-10-19 21:27:46 UTC 2023-05-09 11:59:24 UTC hello@padlet.com kmbeal https://padlet.com/kmbeal/upqoa6tah5e7/wish/399894521 Start by finding the anti-derivative. In this case it is -cos(x)
Then we integrate over our chosen interval: Let's do 0<=x<=pi, 0<=x<=2pi, and pi<=x<=2pi

Geometric approximation
Interval one:
1/2*3.14*1=1.57
Interval two:
1/2*3.14*1-1/2*3.14*1=0
Interval three:
1/2*3.14*(-1)=-1.57

Integrating for our first interval we get -cos(pi)-(-cos(0))=1+1=2
Second interval:
-cos(2pi)-(-cos(0))=-1+1=0
Third interval:
-cos(2pi)-(-cos(pi))=-1-1=-2]]> 2019-10-19 21:37:02 UTC https://padlet.com/kmbeal/upqoa6tah5e7/wish/399894521 kmbeal https://padlet.com/kmbeal/upqoa6tah5e7/wish/399895341 Start by finding the anti-derivative. In this case it is (1/3)x^3
Then we integrate over our chosen intervals: Let's do 0<=x<=2, 2<=x<=5, -2<=x<=2

Geometric approximation
Interval one:
1/2*2*4=4
Interval two:
3*4+1/2*3*21=43.5
Interval three:
See that it's the same as our first interval times 2. 2*4=8

Now calculate the definite integral:
First interval:
1/3(0^3)-(1/3(2^3))=0-8/3=-8/3
Second Interval:
1/3(2^3)-(1/3(5^3))=8/3-125/3=-117/3=-39
Third Interval:
1/3(-2^3)-(1/3(2^3))=-8/3-8/3=-16/3]]> 2019-10-19 21:45:54 UTC https://padlet.com/kmbeal/upqoa6tah5e7/wish/399895341