 Hello and welcome to the session. Let us discuss the following question. Question says using integration, find the area of region bounded by the triangle whose vertices are minus 1, 0, 1, 3 and 3, 2. Let us now start with the solution. First of all let us assume that vertices of triangle are a, b and c. So we can write, let a minus 1, 0, b, 1, 3 and c, 3, 2 with the vertices of triangle a, b, c. Now here we have drawn a diagram to show triangle a, b, c. Now we know equation of line passing through points x1, y1 and x2, y2 is y minus y1 is equal to y2 minus y1 upon x2 minus x1 multiplied by x minus x1. Now using this result we can find equation of line a, b it is equal to y minus 0 is equal to y2 minus y1 that is 3 upon x2 minus x1 that is 2 multiplied by x minus x1 that is minus minus 1. Now simplifying this equation we get y is equal to 3 upon 2 multiplied by x plus 1. We know minus sign multiplied by minus sign is equal to plus sign. Now we will find equation for line b, c. Here we can write y minus y1 and y1 is equal to 3 here. So we can write 3 here is equal to y2 minus y1 that is 2 minus 3 upon x2 minus x1 that is 3 minus 1 multiplied by x minus x1. We know x1 is equal to 1 so here we can write 1. Now simplifying this equation further we get y minus 3 is equal to minus 1 upon 2 multiplied by x minus 1. Now multiplying both the sides of the equation by minus 2 we get minus 2y plus 6 is equal to x minus 1. Now subtracting 6 from both the sides we get minus 2y is equal to x minus 7. Now dividing both the sides by minus 2 we get y is equal to 7 minus x upon 2. Now let us find out equation for line ac. Here y1 is equal to 0. So we will write here y minus y1 that is 0 is equal to y2 that is 2 minus 0 upon x2 minus x1 that is 3 minus minus 1 multiplied by x minus minus 1. We know minus 1 is the x1. Now simplifying this equation we get value of y is equal to x plus 1 upon 2. Now clearly we can see in the figure that area of triangle ABC is equal to area of triangle ABN plus area of trapezium BNMC minus area of triangle ACM. So we can write area of triangle ABC is equal to area of triangle ABN plus area of trapezium BNMC minus area of triangle ACM. Clearly we can see area of triangle ABN is given by the integral from minus 1 to 1 3 upon 2 multiplied by x plus 1 dx. So we can write area of triangle ABN is equal to definite integral from minus 1 to 1 3 upon 2 multiplied by x plus 1 dx. We know total area of this triangle is equal to value of y at this point minus value of y at this point and we know value of y at this point is equal to 3 upon 2 multiplied by x plus 1. We know equation of line AB is this expression and value of y at this point that is n is equal to 0. So we get BN is equal to 3 upon 2 multiplied by x plus 1 minus 0 that is 3 upon 2 multiplied by x plus 1 only. Now let us discuss what is area of trapezium BNMC. It is equal to definite integral from 1 to 3 7 minus x upon 2 dx. If we draw a small vertical strip in this trapezium then area of that vertical strip is equal to 7 minus x upon 2 minus 0. We know value of y at any of the point on this line is equal to 7 minus x upon 2 and value of y on any of the point on line NM is equal to 0. So we get area of this trapezium is equal to this integral. Now we will find out area of triangle ACM. Now clearly we can see it is equal to definite integral from minus 1 to 3. We know coordinate of x at point M is 3. So we can write definite integral from minus 1 to 3 x plus 1 upon 2 dx. We know if we draw any vertical strip in triangle ACM having width equal to dx we get area of that strip is equal to x plus 1 upon 2 multiplied by dx. We know vertical distance is equal to value of y on this line minus value of y on this line. And we know that value of y on AC is equal to x plus 1 upon 2 and value of y on AM is equal to 0. So we get x plus 1 upon 2 multiplied by dx is equal to area of that small vertical strip. So we get total area of triangle ACM is equal to definite integral from minus 1 to 3 x plus 1 upon 2 dx. Now let us find out area of triangle ABC. It is equal to definite integral from minus 1 to 1 3 upon 2 x plus 1 dx plus definite integral from 1 to 3 7 minus x upon 2 dx minus definite integral from minus 1 to 3 x plus 1 upon 2 dx. We know area of triangle ABC is equal to area of triangle ABN plus area of trapezium BNMC minus area of triangle ACM. Now we will solve these integrals. Now this definite integral is equal to 3 upon 2 multiplied by x square upon 2 plus x where lower limit is minus 1 and upper limit is 1. We know 3 upon 2 is constant value. Integral of x is x square upon 2 and integral of 1 is x. Now we will solve this integral. It is equal to 1 upon 2 multiplied by 7x minus x square upon 2 where lower limit is 1 and upper limit is 3. 1 upon 2 is constant here and you know integral of 7 is 7x and integral of x is x square upon 2. Now we will solve this integral. In this integral 1 upon 2 is constant and integral of x is x square upon 2 and integral of 1 is x and lower limit is minus 1 upper limit is 3. Let us now find out values of the integrals at given limits. Now here we can write 3 upon 2 multiplied by 1 square upon 2 plus 1 minus square of minus 1 upon 2 plus minus 1. Now we will apply limits here. We can write 1 upon 2 multiplied by 7 multiplied by 3 minus 3 square upon 2 minus 7 minus 1 square upon 2. Here we can write 7 multiplied by 1. Now we will apply limits here. It is minus 1 upon 2 multiplied by 3 square upon 2 plus 3 minus minus 1 square upon 2 plus minus 1. Now this expression is further equal to 3 upon 2 multiplied by 1 upon 2 plus 1 minus 1 upon 2 plus 1 plus 1 upon 2 multiplied by 21 minus 9 upon 2 minus 7 plus 1 upon 2 minus 1 upon 2 multiplied by 9 upon 2 plus 3 minus 1 upon 2 plus 1. Now here this 1 upon 2 and this 1 upon 2 will get cancelled and we get 3 upon 2 multiplied by 2. Now in this bracket we get 1 upon 2 multiplied by 21 minus 7 is equal to 14 and minus 9 upon 2 plus 1 upon 2 is equal to minus 4. Here we can write minus 1 upon 2 multiplied by 4 plus 4. Now simplifying further we get 3 plus 5 minus 4 which is further equal to 4. So we get area of triangle ABC is equal to 4 square units. So we get required area of triangle ABC is equal to 4 square units. This completes the session. Hope you understood the solution. Take care and keep smiling.