00:01
Hello everyone, the given problem is in the concept of area under the curve.
00:04
In the problem we have to find the area of the given region r which is bounded by the given curves y equal to x squared added to 4 and the straight line given y equal to minus x added to 10.
00:15
It is also given that the area is in the first word.
00:18
Now to solve the problem we first try to find the graphical representation of the given curve that is parabolic curve and the straight line.
00:26
In the graphical representation you can see that the curve with the grid outline it defines the y equal to x squared to 4 and with the blue outline we have the straight line y equal to minus x added to 10 we find the intersecting points in the form of therefore we get x squared added to 4 equal to minus x added to 10 further simplifying we get x squared added to x minus 6 that is equal to 0 we get x plus 3 multiplied with x minus 2 value being equal to 0 that will give us 2 values for 2 and minus 3.
01:03
Now we can see that the minus 3 value x equal to minus 3 that will be in the second quadrant.
01:08
So we don't need to find the value, we just need to check for x equal to 2.
01:12
Now at x equal to 2, the value of y, it will be 8.
01:17
So we find the value 2 8 and the intersecting point for the light, y equal to minus x plus 10 on the y -axis, it will be 0 .10 as we have to find the area in the first quadrant.
01:27
So this area in the yellow shade that we have to find.
01:30
The intersecting points are 0 .10 and 2 .8 so we are finding the area over the x -axis in the point range 0 to 2.
01:39
Also to find the area we first find the area in the interval 0 to 2 for the straight line then we subtract the value or area under the given curve y equal x squared 2 so if we take the area for the given curve as y1 and the area under the straight line as y2 then we can write the total area a in the interval 0 to 2 first we take the area as 1.
02:02
First we take the area as the area the straight line therefore y2 then we take the curve we subtract that so minus y1 t the value will be 0 to 2 for y 2 we have the value minus x added to 10 minus x squared minus 4 that was given for the parabolic curve further simplifying weight for minus 6 we will have minus x squared divided by 2 10 minus 4 therefore 6 integral will give us 6 for minus x square we get x cubed divided by 3 the integral value in 0 to 2 so the limit will be 0 to 2 putting the values we get 2 square minus 0 therefore 4 divided by 2 we get minus 2 6 multiplied with 2 therefore 12 and 2 cube divided by 3 therefore 8 divided by 3 we get the value 10 minus 8 divided by 3 therefore the required area it becomes 22 divided by 3 square units in the second part, we have to find the area of the region that is bounded by the curves y equal to x sq minus 2x and the second one y equal to minus x square added to 4.
03:21
To find the solution, we first find the intersecting points of these two curves and the diagrammatic representation also.
03:28
From the graphical representation we can see that with the green outline we have the curve y equal to x square minus 2x and with the purple outline we have the curve y equal to minus x squared added to 4...