A definite integral is an integral that calculates a fixed value for the area under a curve between two specified limits. The resulting value represents the sum of all infinitesimal quantities within these boundaries. If we integrate any function within a fixed interval, it is called a Definite Integral.
The starting point of the interval is the lower limit of the definite integral, whereas the endpoint of the interval is the upper limit of the definite integral. The definite integral is denoted by \int^{b}_{a}f(x)dx. Here, 'a' and 'b' are the limits of the integral. "a" is called the lower limit of the integral, and "b" is called the upper limit.
Definite IntegralThe definite integral can also be defined as the antiderivative of any function in the interval [a, b], and now for any function F(x) whose derivative is f(x), then in the interval [a, b] its definite integral is defined as,
\int^{b}_{a}f(x)dx = F(b) - F(a)
Note: This result came from the Fundamental Theorem of Calculus.
Definite Integral using Riemann Sum
Definite integral of a function f(x) from a to b using the Riemann sum is defined as the limit of the sum of areas of rectangles under the curve as the number of rectangles becomes infinitely large and their width approaches zero. Formally,
\int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(c_i) \Delta x_i
Where:
- [a, b] is the interval of integration.
- The interval is divided into n small subintervals, each with width Δxi.
- ci is a chosen point within the ith subinterval.
- As n approaches infinity, Δxi becomes very small.
Definite Integral as Limit of a Sum
Assuming that ƒ is a continuous function and positive on the interval [a, b]. So, its graph is above the x-axis.
Definite integral \int^{b}_{a}f(x)dx is the area bounded by the curve y = f(x), the ordinates x = a and x = b and x-axis.
Now to evaluate this area, consider the region ABCD in the figure below,
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Let x0 = a and xn = b.
Now divide the interval [a, b] into n equal subintervals denoted by [x0, x1], [x1, x2], [x2, x3] ....[xr-1, xr] .....[xn-1, xn]
where x0 = a, x1 = a + h, x2 = a + 2h .... and xn = a + nh or n =\frac{b-a}{h}. As n ⇢ ∞, h ⇢ 0.
The region ABCD under consideration is the sum of n subregions, where each subregion is defined on subintervals [xr – 1, xr], where, r = 1, 2, 3, …, n.
It can be seen in the above figure that, now the area of the rectangle PQFR is calculated as,
A = PQ × PR
⇒ A = (xr– xr–1) × f(xr-1)
As xr– xr–1 → 0, i.e., h → 0, the area above becomes a nearly perfect rectangle. Now the area under the curve can be broken into n different rectangles adding all these rectangles' areas we get the area under the curve.
s_{n} = h[f(x_{0}) + .... + f(x_{n})]
= h\sum^{n-1}_{r=0}f(x_{r})
S_{n} = h[f(x_{1}) + .... + f(x_{n})]
= h\sum^{n}_{r=1}f(x_{r})
sn and Sn denote the sum of areas of all lower rectangles and upper rectangles raised over subintervals [xr-1, xr] for r = 1, 2, 3,.... respectively.
As n → ∞ strips become narrower and narrower, so, the limiting values of (2) and (3) are the same in both cases and the common limiting value is the required area under the curve.
So, \lim_{n \to \infty}s_{n}
\\= lim_{n \to \infty}S_{n} \ = \int^{b}_{a}f(x)\ dx
Now, this equation can also be re-written as,
\int^{b}_{a}f(x)dx = \lim_{n \to ∞}[f(a) + f(a + h) + f(a + 2h) + f(a + 3h) ......f(a + (n-1)h)] \\
where, h = \frac{b-a}{n} \to 0 \text{ as } n \to \infty
This expression is knows as definition of definite integral as limit of sum.
Example: Find \int^{2}_{0}(x^{2} + 1)dx as the limit of sum.
Solution:
By the definition given above,
\int^{b}_{a}f(x) = \lim_{h \to 0}[f(x) + f(a + h) + f(a + 2h) + f(a + 3h) ......f(a + (n-1)h)]
Where h = \frac{b-a}{n} \to 0 \text{ as } n \to \infty
Here, a = 0 and b =2, f(x) = x2 + 1, h = \frac{2 - 0}{n} = \frac{2}{n}
\int^{2}_{0}(x^{2} + 1) = 2\lim_{n \to \infty}\frac{1}{n}[f(0) + f(\frac{2}{n}) + f(\frac{4}{n}) + ......f(\frac{2(n-1)}{n})]
= 2\lim_{n \to \infty}\frac{1}{n}[1 + (\frac{2}{n}^{2} + 1) + \frac{4}{n}^{2} + 1 + ......\frac{2(n-1)}{n}^{2} + 1]
= 2\lim_{n \to \infty}\frac{1}{n}[\underbrace{(1 + 1+ .... 1)}_\text{n terms} + \frac{1}{n^{2}}(2^{2} + 4^{2} + ..... (2n - 2)^{2}]
= 2\lim_{n \to \infty}\frac{1}{n}[n + \frac{4}{n^{2}}(1^{2} + 2^{2} + .... (n-1)^{2}]
= 2\lim_{n \to \infty}\frac{1}{n}[n + \frac{4}{n^{2}}\frac{((n-1)n(2n-1)}{6}]
\text{Evaluating above limit} = \frac{14}{3}
How to Find Definite Integral?
Steps for Calculating the definite integral using the Fundamental theorem of calculus are given below,
Step 1: Find the indefinite integral \int f(x)dx . Let's call it F(x). There is no need to keep a constant "C", it will cancel out anyway in the end.
Step 2: Find F(b) - F(a) = [F(x)]^a_b which is the value of this definite integral.
Example: Calculate the integral: \int^{3}_{2}x^{2}dx
Solution:
Solving Indefinite Integral we get,
F(x) = \int x^{2}dx \\= \frac{x^{3}}{3}
Let the value of definite integral be S. and a= 2 and b = 3
S = F(b) - F(a)
⇒ S = (3)3/3 - (2)3/3
⇒ S = 9 - 8/3
⇒ S = 19/3
Properties of Definite Integral
Various Properties of Definite Integral are:
Property 1: This property states that the limits are interchangeable on definite integrals with an extra negative sign.
\int_{a}^{b}f(x)dx =- \int_{b}^{a}f(x)dx
Property 2: This property has limits from a to itself; therefore, the figure is nothing but a point, and there is no area of a point. Therefore, the integration with such limits is always zero.
\int_{a}^{a}f(x)dx=0
Property 3: This property is valid since C is a Constant that can be taken out from the integration easily as it is not included in the function given.
\int_{a}^{b}cf(x)dx=c \int_{a}^{b}f(x)dx
Property 4: This property states that the value of the Integration will remain the same after splitting the function connected with sum or difference.
\int_{a}^{b}[f(x){_{-}^{+}}g(x)]dx= \int_{a}^{b}f(x)dx{_{-}^{+}}\int_{a}^{b}g(x)dx
Property 5: This property states that the variable used to integrate the function does not matter as long as the limits and the value of the function are the same.
\int_{a}^{b}f(x)dx=\int_{a}^{b}f(t)dt
Property 6:
\int_{a}^{b}cdx = c[b-a] where, C is any constant.
Property 7: This property says that if the value of the function is greater than zero, then its integration will also be greater than zero. If f(x) ≥ 0 for a ≤ x ≤ b then,
\int_{a}^{b}f(x)dx ≥ 0.
Property 8: If the value of a function is greater than the value of another function, then the integration of that function will also be greater than the integration of the other function. If f(x) ≥ g(x) for a ≤ x ≤ b then,
\int_{a}^{b}f(x)dx ≥ \int_{a}^{b}g(x)dx
Property 9:
If p ≤ f(x) ≤ P for a ≤ x ≤ b then p[a-b]≤ \int_{a}^{b}f(x)dx≤ P[a-b]
Property 10:
|{\int_{a}^{b}f(x)dx }| ≤ \int_{a}^{b}|f(x)|dx
Integration by Parts 11 : If u=u(x) and dv=v′(x) dx, then
\int_{a}^{b} u(x) \cdot v'(x) \, dx = \left[ u(x) \cdot v(x) \right]_a^b - \int_{a}^{b} v(x) \cdot u'(x) \, dx
Even and Odd Functions 12: An even function satisfies the condition f(-x) = f(x) For an even function, the integral over the symmetric interval [-a, a] and An odd function satisfies the condition f(-x) = -f(x) For an odd function, the integral over the symmetric interval [-a, a] is
\int_{-a}^{a} f(x) \, dx = 2 \int_0^a f(x) \, dx ( f(x) is even function )
\int_{-a}^{a} f(x) \, dx = 0 ( f(x) is odd function )
Applications of Definite Integral
Definite integral finds its application in various fields, some of the important Applications of Definite Integral are,
- Finding areas of various curves such as circles, parabolas, ellipses, and others.
- Solving the infinite series and finding their sum.
- Finding Divergence and curl of the vectors and others.
Note on Continuity for Definite Integrals:
- For a definite integral \int_{a}^{b} f(x) \, dx , the function f(x) must be continuous on the closed interval [a,b]
- If f(x) has a removable discontinuity (e.g., a hole), the integral can still be evaluated.
- If f(x) has an infinite discontinuity (e.g., a vertical asymptote), the integral may require improper integration to handle the limits.
- Discontinuous functions might still be integrable depending on the type of discontinuity and the techniques used.
Definite Integral Examples
Example 1: Evaluate, \int_{0}^{\pi/2}cosx dx
Solution:
\int_{0}^{\pi/2}cosx dx= [sinx]_{0}^{\pi/2}\\(sin\pi/2)-(sin0)=1-0=1
Example 2: Find the Integral: \int_{0}^{\frac{\pi}{6}}cos\ 3x\ dx
Solution:
Solving, \int_{0}^{\frac{\pi}{6}}cos\ 3x\ dx=[\frac{sin3x}{3}]_{0}^{\frac{\pi}{6}}\\\frac{1}{3}[sin(\frac{\pi}{2})-sin0)]=\frac{1}{3}.
Example 3: Integrate \int_{0}^{2}e^xdx
Solution:
\int_{0}^{2}e^xdx=(2-0)lim_{n\to\infty}\frac{1}{n}[e^0+e^\frac{2}{n}+e^\frac{4}{n}+....+e^\frac{2n-2}{n}
Applying G.P, a=1. r=e2/n, we will get,
\int_{0}^{2}e^xdx=(2)lim_{n\to\infty}\frac{1}{n}[\frac{e^\frac{2n}{n}-1}{e^\frac{2}{n}-1}]\\2lim_{n\to\infty}\frac{1}{n}[\frac{e^2-1}{e^\frac{2}{n}-1}
Using,lim_{h\to0}\frac{e^h-1}{h}=1
\int_{0}^{2}e^xdx=e^2-1
Example 4: Evaluate\int_{1}^{2}logxdx
Solution:
\int_{1}^{2}logx.1dx=[logx.x]_{1}^{2}-\int_{1}^{2}\frac{1}{x}.xdx\\
2log2-2+1=(2log2-1)
Example 5: Evaluate \int_{-1}^{2}5^xdx
Solution:
a=-1, b=2, f(x)=5x
h=(b-a)/n=3/n, nh=3
By definition: \int^{b}_{a}f(x) = \lim_{h \to 0}[f(x) + f(a + h) + f(a + 2h) + f(a + 3h) ......f(a + (n-1)h)] \\
\int_{-1}^{2}5^xdx=\lim_{h\to0}h[f(-1)+f(-1+h)+f(-2+h)+.....f(-1+(n-1)h)]\\\lim_{h\to0}h[5^{-1}+5^{-1}.5^h+.....5^{-1}.5^{(n-1)h}\\\lim_{h\to0}h[\frac{5^-1.(5^{nh}-1)}{5^h-1}]\\\lim_{h\to0}\frac{h}{5^h-1}5^{-1}(5^3-1)
\frac{1}{log5}[5^2-\frac{1}{5}]=\frac{124}{5log5}
Definite Integral Practice Problems
Problem 1: Evaluate the definite integral: \int_{0}^{1} (2x^2 + 3x + 1) \, dx
Problem 2: Compute the definite integral: \int_{-2}^{2} x^3 \, dx
Problem 3: Find the value of the integral: \int_{0}^{\pi} \sin(x) \, dx
Problem 4: Calculate the definite integral: \int_{1}^{e} \frac{1}{x} \, dx
Problem 5: Determine the integral: \int_{0}^{2} (4x - x^2) \, dx
Problem 6: Evaluate the integral \int_0^1 x \cdot e^x \, dx
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