What is the Upper Limit of a Sum in Integration?
Consider a curve that lies above the x-axis. This graph’s function is a continuous function with all of its values being non-negative, defined on the closed interval [a, b]. The definite integral ab f(x) dx of any such continuous function “f” is the region bound between the curve, the points “x = a” and “x = b,” and the x-axis.
You can also calculate the sum of nth terms of a given series onlne by using limit of the sum calculator.
Take a look at the graph below:
Let’s assess the area PRSQP between the x-axis, the curve y = f(x), and the coordinates “x = a” and “x = b” in order to comprehend this. Divide the range [a, b] into n equal sub ranges, indicated as [x0, x1], [x1, x2], [x2, x3],…. [xn – 1, xn], where x0 = a, x1 = a + h, x2 = a + 2h, x3 = a + 3h,….. xr = a + rh, and xn = b
Also, n = (b – a)/h. Keep in mind that h = 0 as n =.
Now, the region PRSQP under examination is made up of ‘n’ sub-regions, each of which is defined by subintervals [xr – 1, xr], r = 1, 2, 3,… n. Check out the region ABDM in the illustration above. We are able to note the following:
Areas of the region, the rectangle, and the rectangle (ABLC, ABDCA, ABDM) (1)
Also take note of the fact that all three of these areas approach equality as h or xr – xr – 1 approaches zero. So, sn = h [f(x0) + f(x1) + f(x2) +…. f(xn – 1)] = h r = 0 n 1 f(xr) (2)
Additionally, Sn = h [f(x1) + f(x2) +… + f(xn)] = h r = 1n f (xr) (3)
For r = 1, 2, 3,…, n, respectively, sn and Sn represent the total of all lower rectangles and upper rectangles raised over subintervals [xr-1, xr]. Equation (1) can be rewritten as: sn area of the region (PRSQP) Sn to put things into perspective. (4)
Read more Issues with Integrals
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These strips become thinner as n increases.
Furthermore, it is presumable that the limiting values of (2) and (3) are the same in both situations, with the needed area under the curve serving as the common limiting value. In a symbolic sense, we have limn. S = limn = PRSQP: sn = area of the region = ab f(x) dx (5) Any area that lies between those of the rectangles below the curve and those of the rectangles above the curve has this area as its limiting value. We will use the rectangles with heights equal to the curve at the left-hand edge of each sub-interval for convenience. As a result, equation (%) is changed to: ab f(x) dx = limn h [f(a) + f(a + h) +…. + f(a + n – 1h)].
Alternatively, f(a) + f(a + h) +…. + f(a + n – 1h) [ab f(x) dx = (b – a) limn (1/n)] … (6) (6)
In this case, h = (b – a)/n 0 as n . The limit of a sum is defined by this equation as the Definite Integral.
It should be noted that the value of the definite integral of a function over a given interval depends on the interval and the function, not the integration variable we select to represent the independent variable. Therefore, the integration variable is often referred to as a dummy variable.
Figure 1
Find the limit of a sum as 02 (x2 + 1) dx.
Solution: Knowing that ab f(x) dx = (b – a) limn (1/n) [f(a) + f(a + h) +…. + f(a + n – 1h)] from equation (6) above.
h = (b – a)/n where
A = 0, B = 2, f(x) = (x2 + 1), and h = (2 – 0)/n = 2/n are the values in this example. The result is that 02 (x2 + 1) dx = 2 limn (1/n) [f(0) + f(2/n) + f(4/n) +…. + f(2n – 1/n)] [1 + {(22/n2) + 1} + {(42/n2) + 1} + …. + {(2n – 2)2/n2 + 1}]
= 2 limn → ∞ (1/n) [1+1+1+1+…+1(n-times)] + 1/n2 [22 + 42 + … (2n – 2)2]
= 2 limn → ∞ (1/n) [n + 22/n2 (12 + 22 + … (n – 1)2]
= 2 limn → ∞ (1/n) [n + 4/n2 {(n – 1) n (2n – 1) / 6}]
= 2 limn → ∞ (1/n) [n + 2/3 {(n – 1) (2n – 1) / n}]
= 2 limn → ∞ (1/n) [n + 2/3 (1 – 1/n) (2 – 1/n)]
As n → ∞, 1/n → 0. Thus, 02 (x2 + 1) dx = 2 [1 + 4/3] = 14/3 is the result.
Additional Examples Solved for You
Determine the limit of a sum using ab x dx.
Answer: Knowing that ab f(x) dx = (b – a) limn (1/n) [f(a) + f(a + h) +…. + f(a + n – 1h)] from equation (6) above.
h = (b – a)/n where
In this illustration, a = a, b = b, f(x) = x, and h = (b – a)/n are all true.
Aside from that, f(a) = a, f(a + h) = a + h, and f(a + 2h) = a + 2h.
f(a + 3h) = a + 3h ……
A + (n – 1)h = f(a + (n – 1)h
As a result, ab x dx = (b – a) limn ab x dx = (1/n) [f(a) + f(a + h) +… + f(a + n – 1h)]
= (1/n) limn (b – a) [a+ (a+h) + (a+2h) +…. + (a+(n-1)h)]
= (1/n) limn (b – a) [a+ a+ a+ a+… + a(n-times)] + h + 2h + …. + (n – 1)h}
= (1/n) limn (b – a) [na + h(1 + 2 + ….. + (n – 1)]
Now that we are aware of it, n(n + 1)/2 = 1 + 2 + 3… + n. In light of this, 1 + 2 + 3 +… + (n – 1) = (n – 1)(n – 1 + 1)/2 = n(n – 1)/2
As a result, ab x dx = (b – a) limn ab x dx = (1/n) [na + hn(n – 1)/2]
= (b – a) limn [hn(n – 1)/2n + na/n]
= [a + (n – 1)h/2] (b – a) limn
We obtain ab x dx = (b – a) limn by substituting (b – a)/n for h. [a + (n – 1) (b – a) / 2n]
[a + (n/n – 1/n)] = (b – a) limn {(b – a)/2}]
= limn = (b – a) [a + (1 – 1/n) {(b – a)/2}]
= (b – a) (b – a) [a + (1 – 1/∞) {(b – a)/2}]
= (b – a) (b – a) [a + (1 – 0){(b – a)/2}]
= (b – a) (b – a) [a + (b – a)/2] equals (b – a); (b + a)/2 equals (b2 – a2)/2;
As a result, [(b2 – a2)/2] is the limit of the definite integral ab x dx.
Define a definite integral?
In order to determine the net area between a function and the x-axis, we can define it as an exact limit and summation that we looked at in the previous section. Additionally, see how closely the notation for definite integrals resembles that for indefinite integrals.
Why do we employ integrals?
To find volumes, areas, central points, and a variety of other helpful things, we employ integrals. However, we frequently use it to determine the region under a function’s graph, such as this one: By combining slices that are close to zero in width, we can locate this region. Additionally, there are integration guidelines that aid in obtaining the solution.
What are the definitions of definite and indefinite integral?
The solution is a particular area, and the term “definite integral” refers to an integral with integration bounds. The indefinite integral, on the other hand, yields a function of the independent variable(s).
Can a definite integral be negative?
Since an integral measures the distance between the x-axis and the curve over a given period of time, it is possible for an integral to be negative. Additionally, the outcome is good if all of the interval’s areas are above the x-axis but below the curve.
