Which equation has no solution?
1 answer:
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Answer:
Step-by-step explanation:
I'm assuming you meant to type in
because you can only have removable discontinuities where there is a rational (fraction) function. Begin by factoring both the numerator and denominator to
and cancelling out like terms would have us eliminating the (x + 3). That is where there is a removable discontinuity. It leaves a hole. The other discontinuity, (x + 1) doesn't cancel out so it is a non-removable discontuinity, which is a vertical asymptote.
The removable discontinuity is at -3. There is no y value at x = -3 (remember there's only a hole here), because -3 causes the denominator to go to 0 and we all know that having a 0 in the denominator of a fraction is a big no-no!!!
The expression of integral as a limit of Riemann sums of given integral is 4
∑
from i=1 to i=n.
Given an integral .
We are required to express the integral as a limit of Riemann sums.
An integral basically assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinite data.
A Riemann sum is basically a certain kind of approximation of an integral by a finite sum.
Using Riemann sums, we have :
=
∑f(a+iΔx)Δx ,here Δx=(b-a)/n
=f(x)=
⇒Δx=(5-1)/n=4/n
f(a+iΔx)=f(1+4i/n)
f(1+4i/n)=
∑f(a+iΔx)Δx=
∑
=4∑
Hence the expression of integral as a limit of Riemann sums of given integral is 4
∑
from i=1 to i=n.
Learn more about integral at brainly.com/question/27419605
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