
John M. answered 03/22/20
Math Teacher/Tutor/Engineer - Your Home, Library, MainStreet or Online
1,
∫baf(x)dx≈Δx(f(x0)+f(x1)+2f(x2)+...+f(xn−2)+f(xn−1)), where Δx=b−an
.
We have that a=0
, b=2, n=4
.
Therefore, Δx=(2−0)/4=1/2
.
Divide the interval [0,2]
into n=4 subintervals of length Δx=12: a=[0,12],[12,1],[1,32],[32,2]=b
Now, we just evaluate the function at the left endpoints:
f(x0)=f(a)=f(0)=3=3
f(x1)=f(12)=72=3.5
f(x2)=f(1)=4=4
f(x3)=f(32)=92=4.5
Finally, just sum up the above values and multiply by Δx=12
: 12(3+3.5+4+4.5)=7.5Answer: 7.5.
2.
∫baf(x)dx≈Δx(f(x0)+f(x1)+2f(x2)+...+f(xn−2)+f(xn−1)), where Δx=b−an
.
We have that a=0
, b=2, n=4
.
Therefore, Δx=(2−0/)4=1/2
.
Divide the interval [0,2]
into n=4 subintervals of length Δx=12: a=[0,12],[12,1],[1,32],[32,2]=b
.
Now, we just evaluate the function at the left endpoints:
f(x0)=f(a)=f(0)=2=2
f(x1)=f(12)=72=3.5
f(x2)=f(1)=5=5
f(x3)=f(32)=132=6.5
Finally, just sum up the above values and multiply by Δx=12
: 12(2+3.5+5+6.5)=8.5Answer: 8.5.
3.
∫baf(x)dx≈Δx(f(x0)+f(x1)+2f(x2)+...+f(xn−2)+f(xn−1)), where Δx=b−an
.
We have that a=0
, b=5, n=4
.
Therefore, Δx=(5−0)/4=5/4
.
Divide the interval [0,5]
into n=4 subintervals of length Δx=54: a=[0,54],[54,52],[52,154],[154,5]=b
.
Now, we just evaluate the function at the left endpoints:
f(x0)=f(a)=f(0)=0=0
f(x1)=f(54)=54=1.25
f(x2)=f(52)=52=2.5
f(x3)=f(154)=154=3.75
Finally, just sum up the above values and multiply by Δx=54
: 54(0+1.25+2.5+3.75)=9.375Answer: 9.375.
4.
∫baf(x)dx≈Δx(f(x0)+f(x1)+2f(x2)+...+f(xn−2)+f(xn−1)), where Δx=b−an
.
We have that a=2
, b=5, n=4
.
Therefore, Δx=(5−2)/4=3/4
.
Divide the interval [2,5]
into n=4 subintervals of length Δx=34: a=[2,114],[114,72],[72,174],[174,5]=b
.
Now, we just evaluate the function at the left endpoints:
f(x0)=f(a)=f(2)=2=2
f(x1)=f(114)=114=2.75
f(x2)=f(72)=72=3.5
f(x3)=f(174)=174=4.25
Finally, just sum up the above values and multiply by Δx=34
: 34(2+2.75+3.5+4.25)=9.375Answer: 9.375.