 Hello and welcome to the session. In this session we will discuss about the definite integral. A definite integral is denoted by integral a to b fx dx where this a is called the lower limit of the integral and b is called the upper limit of the integral. This definite integral has a unique value. The definite integral is also introduced as a limit of sum. Let's see how do we do this. Integral a to b fx dx is equal to limit h tends to 0 h into f of a plus f of a plus h plus and so on up to f of a plus n minus 1 into h. Or this can also be written in the form integral a to b fx dx is equal to b minus a limit h tends to infinity 1 upon n into f of a plus f of a plus h plus and so on up to f of a plus n minus 1 into h where we have h is equal to b minus a upon n h tends to 0 as n tends to infinity. This expression is known as the definition of definite integral as the limit of sum. Let us try and evaluate the integral a to b x dx as limit of sums. Here as you can see the lower limit is a upper limit is b and the function fx is equal to x and we know that h is equal to b minus a upon n. Now using the definition of the definite integral as a limit of sum this can be written as integral a to b x dx is equal to b minus a limit n tends to infinity 1 upon n f of a that would be a plus f of a plus h that is a plus h plus f of a plus 2h that is a plus 2h plus and so on up to f of a plus n minus 1 into h that is a plus n minus 1 into h. This is equal to b minus a into limit n tends to infinity 1 upon n into n a plus h into 1 plus 2 plus and so on up to n minus 1. This becomes equal to b minus a into limit n tends to infinity 1 upon n into n a plus h into now we know that sum of n terms is n into n plus 1 upon 2. So sum of n minus 1 terms would be n into n minus 1 upon 2. So here we have n minus 1 into n upon 2. This is equal to b minus a into limit n tends to infinity multiplied by a plus now the value for h is b minus a upon n into n minus 1 upon 2. This is equal to b minus a into limit n tends to infinity a plus b minus a upon 2 limit n tends to infinity n minus 1 upon n. This gives us b minus a into b plus a upon 2 which is equal to b square minus a square upon 2. So the definite integral a to b x dx is written as b square minus a square upon 2 as the limit of sum. Now before discussing the fundamental theorem of calculus, firstly let's see what is the area function. We have defined the integral a to b fx dx as the area of the region bounded by the curve y is equal to fx where x is greater than equal to a and less than equal to b the x axis and the ordinate x equal to a and x equal to b. That is this shaded region is the integral a to b fx dx. Now if we have a point x in a b that is this point then the integral a to x fx dx is the area function and it is denoted by ax. That is this shaded region is the area function ax that is integral a to x fx dx. Based on this definition of area function we have two basic fundamental theorems of integral calculus. First let's see what is the first fundamental theorem of integral calculus. According to this we have let the area function be defined by ax is equal to integral a to x fx dx for all x greater than equal to a where the function f is assumed to be continuous on close interval a b then a dash x is equal to fx for all x belongs to close interval a b. Next we see the second fundamental theorem of integral calculus according to which we have let f be a continuous function of x defined on the close interval a b and let capital F be another function such that dy dx of capital Fx is equal to small fx for all x in the domain of f. Then we have integral a to b fx dx is equal to capital Fx plus c from limits a to b equal to fb minus fa. That is integral a to b fx dx is equal to value of nt derivative capital F of small f at the upper limit b minus value of same nt derivative at the lower limit a. This theorem is very useful to calculate the definite integral more easily without calculating the limit of a sum. Let's try and find out the value of the definite integral i equal to integral limits 2 to 3 x cube dx using the second fundamental theorem of integral calculus. Here capital Fx would be given by integral x cube dx which is equal to x to the power 4 upon 4. Here we have a is equal to 2 and b is equal to 3. So by second fundamental theorem of integral calculus we get i is equal to f of 3 minus f of 2 that is equal to 3 to the power 4 upon 4 minus 2 to the power 4 upon 4 which is equal to 81 upon 4 minus 16 upon 4 and this is equal to 65 upon 4. This is the value for the given integral i. So this completes the session. Hope you have understood what is a definite integral, how we evaluate the definite integral as a limit of sum, what is an area function and the two fundamental theorems of integral calculus and how do we evaluate the definite integral using second fundamental theorem of integral calculus.