(a) Determine Δx and xi. Δx= xi= (b) Using the definition mentioned above, evaluate the integral. Value of integral ?

(c) evaluate the integral. Value of integral ?

Answer 1

(a) #Delta x = 3/n#, and #x_i=1+(3i)/n#

(b) # int_(1)^4 \ x^2+2x-5 \ dx = 21 #

By definition of an integral, then, using Riemann sums

# int_a^b \ f(x) \ dx #
represents the area under the curve #y=f(x)# between #x=a# and #x=b#. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles. The limit of which is known as the Riemann sum (or the limit definition of an integral)

That is

# int_a^b \ f(x) \ dx = lim_(n rarr oo) (b-a)/n sum_(i=1)^n \ f(a + i(b-a)/n)#
And we partition the interval #[a,b]# equally spaced using:
# Delta = {a+0((b-a)/n), a+1((b-a)/n), ..., a+n((b-a)/n) } #
# \ \ \ = {a, a+1((b-a)/n),a+2((b-a)/n), ... ,b-a } #
Here we have #f(x)=x^2+2x-5# and so we partition the interval #[1,4]# using:
# Delta = {1, 1+1(3)/n, 1+2 (3)/n, 1+3 (3)/n, ..., 4 } #

And so:

# I = int_(1)^4 \ (x^2+2x-5) \ dx #
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(1+(3i)/n) #

Providing for us:

#Delta x = 3/n#, and #x_i=1+(3i)/n#

By applying the meaning of f(x), we obtain:

# I = lim_(n rarr oo) 3/n sum_(i=1)^n \ {(1+(3i)/n)^2+2(1+(3i)/n)-5} #
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {1+(6i)/n+(9i^2)/n^2 +2+(6i)/n-5}#
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {(9i^2)/n^2 +(12i)/n-2}#
# \ \ = lim_(n rarr oo) 3/n {9/n^2 sum_(i=1)^n i^2+ 12/n sum_(i=1)^n i-sum_(i=1)^n2}#

Applying the common formula for summation:

# sum_(r=1)^n r \ = 1/2n(n+1) # # sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) #

Thus, we have:

# I = lim_(n rarr oo) 3/n {9/n^2 1/6n(n+1)(2n+1)+ 12/n 1/2n(n+1)-2n}#
# \ \ = lim_(n rarr oo) 3/n {3/(2n)(n+1)(2n+1)+ 6(n+1)-2n}#
# \ \ = lim_(n rarr oo) 3/(2n^2) {3(2n^2+3n+1)+ 12n(n+1)-4n^2}#
# \ \ = lim_(n rarr oo) 3/(2n^2) {6n^2+9n+3+ 12n^2+12n-4n^2}#
# \ \ = 3/2 lim_(n rarr oo) 1/n^2 {14n^2+21+3}#
# \ \ = 3/2 lim_(n rarr oo) {14+21/n+3/n^2}#
# \ \ = 3/2 {14+0+0}#
# \ \ = 21#

Applying Calculus

Utilizing Calculus and our understanding of integration to determine the solution, in contrast, yields:

# I = int_(1)^4 \ x^2+2x-5 \ dx # # \ \ = [x^3/3+x^2-5x]_(1)^4 # # \ \ = (64/3+16-20)-(1/3+1-5) # # \ \ = 64/3-4 - 1/3+4 # # \ \ = 21 #
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Answer 2

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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