 Hello and welcome to the session. In this session we discussed the following question which says using integration find the area of the region bounded by the parabola y square equal to 4x and the circle 4x square plus 4y square equal to 9. Before proceeding on to the solution let's see what is the area of the region bounded by the curve y equal to fx x axis and the lines x equal to a and x equal to b where we have b is greater than a is given by area equal to integral a to b y dx or we can also write this as integral a to b fx dx. This is the key idea for this question. Now we move on to the solution. Now we are given the equation of the circle as 4x square plus 4y square equal to 9 that is we have x square plus y square equal to 9 upon 4 or we can also say that x square plus y square is equal to 3 upon 2 the whole square. So this equation represents a circle with center O with coordinate 00 and radius 3 upon 2. Let this equation of the circle be equation 1. Now we have equation of the parabola given to us is y square equal to 4x let this be equation 2. Now we find the points of intersection of equation 1 and 2. So for this we substitute y square equal to 4x in 1. So we get x square plus 4x is equal to 9 upon 4 that is 4x square plus 16x minus 9 equal to 0. So from here we have 4x square plus 18x minus 2x minus 9 equal to 0. From here we get 2x into 2x plus 9 minus 2x plus 9 equal to 0 that is we have 2x minus 1 into 2x plus 9 equal to 0 which gives us that either 2x plus 9 equal to 0 or 2x minus 1 equal to 0 which shows that either x equal to minus 9 over 2 or x equal to 1 over 2. Now for x equal to minus 9 upon 2 in y square equal to 4x we get imaginary value for y. So we reject x equal to minus 9 upon 2. Let's find out the value for y when we put x equal to 1 upon 2. So for x equal to 1 upon 2 in y square equal to 4x we get y square equal to 4 into 1 upon 2 that is equal to 2. So this gives us y is equal to plus minus root 2. Thus we get the two curves 1 and 2 intersect at say point A which coordinates 1 over 2 root 2 and a point B which coordinates 1 over 2 minus root 2. Let's have a look at this figure. This is the parabola y square equal to 4x and this is the circle 4x square plus 4y square equal to 9 where O with coordinate 00 is the center of the circle and the radius is 3 upon 2. So point of intersection of the curves that is the parabola and the circle are points A and B. A has coordinates 1 upon 2 root 2 and B has coordinates 1 upon 2 minus root 2. We have the intersection of the circle with the x axis as the point C which coordinates 3 upon 2 0. Now this shaded region is the region bounded by the two curves and we need to find the area of this shaded region that is we have to find the area of O B C A O. Thus we have required area of the enclosed region O B C A O between the two curves would be equal to 2 times the area of the region O D C A O that is equal to 2 times the area of region O D C A O. Now from this figure you can see that the region O D C A O is equal to the region O D A O plus D C A D. So this would be further equal to 2 times the area of the region O D A O plus area of the region D C A D. Now area of the region O D A O would be the area of the region bounded by the curve y square equal to 4x the x axis and the lines x equal to 0 and x equal to 1 over 2. So using the key idea we can write this as this would be equal to 2 times integral 0 to 1 upon 2 y dx the value of y would be given from y square equal to 4x and this would be equal to square root 4x. So here we have integral 0 to 1 upon 2 square root 4x dx plus the area of the region D C A D. Now area of this region D C A D is the area of the region bounded by the curve that is the circle 4x square plus 4y square equal to 9. The x axis and the lines x equal to 1 over 2 and x equal to 3 upon 2 and from this equation of the circle that is 4x square plus 4y square equal to 9 we get the value of y as square root of 9 upon 4 minus x square or we can also say square root of 3 upon 2 whole square minus x square. So now using the key idea we can say that the area of the region D C A D is given by integral 1 upon 2 to 3 upon 2 square root of 3 upon 2 whole square minus x square dx. Now this would be equal to 2 times 2 into x to the power 3 upon 2 over 3 upon 2 and the limits goes from 0 to 1 upon 2 that is we have integrated square root x plus now for this integral we will apply the formula that is integral of square root A square minus x square dx is equal to x upon 2 square root A square minus x square plus A square upon 2 sin inverse x upon A plus C. Here we have A would be equal to 3 upon 2 so we substitute A equal to 3 upon 2 in this case in this formula so we get x upon 2 square root 3 upon 2 whole square minus x square plus 3 upon 2 whole square upon 2 sin inverse x upon 3 upon 2 and the limit here goes from 1 upon 2 to 3 upon 2. Now let's put the limits so this would be equal to 2 times 4 upon 3 into 1 upon 2 to the power 3 by 2 plus 3 by 4 into square root of 3 upon 2 whole square minus 3 upon 2 whole square plus 9 upon 8 sin inverse 1 minus 1 upon 4 into square root of 3 upon 2 whole square minus 1 upon 2 whole square minus 9 upon 8 sin inverse 1 upon 2 upon 3 upon 2. This comes out to be equal to 2 times 4 upon 3 into 1 upon 2 root 2 plus 0 plus 9 upon 8 into sin inverse 1 that is pi upon 2 minus 1 upon 2 root 2 minus 9 upon 8 sin inverse 1 upon 3. This would be further equal to 2 4 upon 3 root 2 plus 9 pi upon 8 minus 1 upon root 2 minus 9 upon 4 sin inverse 1 upon 3. This is equal to 1 upon 3 root 2 plus 9 pi by 8 minus 9 upon 4 sin inverse 1 upon 3 or we could also write this as root 2 upon 6 plus 9 pi by 8 minus 9 upon 4 sin inverse 1 upon 3. So thus the final answer is required area of the region pounded by the two given curves is root 2 upon 6 plus 9 pi by 8 minus 9 upon 4 sin inverse 1 upon 3. So this completes the session hope you have understood the solutions for this question.