 Hi and welcome to the session. Let us discuss the following question. Question says, using integration, find the area of the triangular region whose sides have equations y is equal to 2x plus 1, y is equal to 3x plus 1 and x is equal to 4. First of all, let us understand that area between two curves or we can say area enclosed by two curves is equal to definite integral from a to b fx minus gx dx. Now in this formula a and b are points on x-axis where these two curves fx and gx intersect each other and fx minus dx multiplied by dx is the area of the strip drawn vertically in the area enclosed between the two curves. fx minus dx is the length of the strip and dx is the width of the strip. So total area enclosed between the two curves is given by this formula. So we will use this formula as our key idea to solve the given question. Let us now start with the solution. Now we are given three equations of line that is y is equal to 2x plus 1, y is equal to 3x plus 1 and x is equal to 4. Now let us assume that abc be the triangular region enclosed between these given sides. So we can write that abc be the triangular region enclosed by the given sides. Let us assume that this is the equation for line ab or we can say site ab. So we can write equation of site ab is y is equal to 2x plus 1. Equation of site ac is equal to y is equal to 3x plus 1. Equation of site bc is x is equal to 4. Let us now name these equations as 1, 2 and 3. Solving 1 and 2 we get coordinates of A as 0, 1. Similarly solving 1 and 3 we get coordinates of B as 4, 9. Now solving 2 and 3 we get coordinates of point C and coordinates of C are 4, 13. Now plotting the points A, B and C on the graph we get triangle abc whose area is to be determined. Clearly we can see area of triangle abc is equal to area of trapezium oacm minus area of trapezium oabm. So now we can write area of triangle abc is equal to area of trapezium oacm minus area of trapezium oabm. Now area of trapezium oacm is equal to definite integral from 0 to 4. We know coordinate of x at this point is 0 and at this point is 4. So we can write definite integral from 0 to 4, 3x plus 1 dx. If we draw some vertical strip in trapezium oacm such that its width is equal to dx. Then area of that vertical strip is equal to 3x plus 1 multiplied by dx. We know length of the strip is equal to y of this line minus y of this line, y of this line is 3x plus 1 and y of this line is 0. So we get length of the strip is equal to 3x plus 1 and width of the strip is dx. And this whole area of trapezium is equal to definite integral from 0 to 4 3x plus 1 dx. Similarly we can find area of trapezium oabm. It is equal to definite integral from 0 to 4 2x plus 1 dx. Halfly we can see limits of these two integrals are same. So we can combine these two integrals. So we can write these two integrals as definite integral from 0 to 4 3x plus 1 minus 2x plus 1 dx. Now this is further equal to definite integral from 0 to 4 x dx. Simplifying this bracket we get x. Now this integral is equal to x square upon 2 and limits of the integral are 0 to 4. To find the integral of x dx we have used the formula integral of x raise to the power n dx is equal to x raise to the power n plus 1 upon n plus 1. Now value of x square upon 2 at 4 is equal to 4 square upon 2 and value of x square upon 2 at 0 is equal to 0 square upon 2. Now this term will become 0 and we get 16 upon 2 minus 0 which is further equal to 16 upon 2. Now we will cancel common factor 2 from numerator and denominator both and we get 8 square units. So required area of this triangle is equal to 8 square units. This completes the session. Hope you understood the solution. Take care and have a nice day.