# Definite Integrals

Fundamental Theorem of Calculus: It states that the sum of small changes in any quantity over a period of time adds up to the total quantity.

In the below figure I have shown the function F(x) and I need to find the area of the above curve I take the bounded values of the curve a and b start summing up taking the small values of dx considering a small rectangle and extending the same area throughout the curve.

This is taken as the integral for the whole area under the curve between the limits a and b.

So fundamental theorem takes the form

$\int_{a}^{b}(F(x)dx$ = F(b) - F(a) to find the area under a curve.

F(x) is the integral of f(x);

F(b) is the value of the integral at the upper limit, x=b; and

F(a) is the value of the integral at the lower limit, x=a.

This expression is called a definite integral.

So it clearly says the area can be found out using the above integral and is also known as definite integral.

## Applications of Definite Integrals

Its usage can be seen in a vast number of applications such as finding the volume of solids of revolution, area enclosed between two curves and integral as sum of limit.

## Properties of Definite Integrals

Consider 2 functions f(x) and g(x) which are continues functions then

$\int_{a}^{b} (f(x)+g(x))=\int_{a}^{b}f(x) + \int_{a}^{b} g(x)$

$\int_{a}^{a}f(x)=0$

$\int_{a}^{b}f(x) = \int_{a}^{d}f(x)+\int_{d}^{b}f(x)$

$\int_{b}^{a}f(x)= -\int_{a}^{b}f(x)$

For arbitrary numbers a and b and d belongs to [a, b].

## Definite Integrals Solved Problems

Example 1: Find the integral and then write the upper and lower limits with square brackets, as follows:

$\int_{2}^{3}(5x^{4})+2x+6$

[x 5 + $x^2$ + 6x] 2, 3

The upper and lower limits are written like this to mean they will be substituted into the expression in brackets.

Next, substitute 5 (the upper limit) into the integral:

[(2-3)5 + (2-3) 2 + 6(2-3)]= -1 +1+ 6(-1) = -6

Example 2: Find the integral $\int_{1}^{2}(5+7x^{6})$

Solution : $\int_{1}^{2}(5+7x^{6})$

5x + x7

$5(2-1) + (2-1)^7$

5 + 1 = 6

Example 3: Find the work done if a force F(x) = 5x - 3 N is acting on an object and moves it from x = 1 to x = 6.

Solution: $\int_{1}^{6}(5x-3)$

[$5x^2 - 3x$]

$5(6-1)^2 - 3(6-2)$

125 - 12

113 joules.

Example 4: The expression, v, for velocity in terms of t, the time, we can find the displacement S of a moving object from time t = 0 to time t = 6 by integration is given by $\int_{0}^{6}(v dt)$

If V is given by t2.

Then [$\frac{t^3}{3}$] (After integrating)

216/18 unit/sec.

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