100PTS Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
y=1/x, y=1/x2,x=6

i tried to use the antiderivative but when i plug 0 in, it turns out ln(6)+1/6-ln(0)-1/0 and i dont know what to do with 1/0 :(

Step-by-step explanation:

We'll integrate with respect to x.

Recall that the area between some and some on the interval is given by:

The function will be the upper bounds of our definite integral. The lower bounds will be the intersection of and . Set the functions equal to each other and solve for x:

Therefore, we'll be integrating the area between and on the interval . In integral notation, this is:

To evaluate this integral, recall that and .

Therefore,

Solving yields:

step-by-step explanation:

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )