# Set up the integrals to evaluate the area of the planar region bounded by (y^2)-8y+13, y-x= 9, y = 2 and y = 7 by horizontal slicing, by vertical slicing. Evaluate only for the method that...

Set up the integrals to evaluate the area of the planar region bounded by (y^2)-8y+13, y-x= 9, y = 2 and y = 7

by horizontal slicing, by vertical slicing.

Evaluate only for the method that yields the fewer number of integrals. Sketch the region.

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### 2 Answers

Notice that the curve `y^2 - 8y + 13` falls above the line `y-x = 9` , if `y in [2,7]` , hence, you need to evaluate the following definite integral to find the area bounded by the given curve and lines such that:

`int_2^7 (y^2- 8y + 13 - y + 9)dy = int_2^7 (y^2- 9y+ 22)dy`

Using the property of linearity of integrals yields:

`int_2^7 (y^2 -9y +22)dy = int_2^7 (y^2)dy- int_2^7(9y)dy + int_2^7 (22)dy`

`int_2^7 (y^2 - 9y + 22)dy = (y^3/3- 9y^2/2+ 22y)|_2^7`

`int_2^7 (y^2- 9y+ 22)dy = 343/3 - 8/3- 441/2+ 36/2+ 154- 44`

`int_2^7 (y^2- 9y+ 22)dy = (686- 16- 1323 + 108+ 924- 264)/6`

`int_2^7 (y^2- 9y+ 22)dy = 115/6`

**Hence, evaluating the area bounded by the given curve and lines yields `int_2^7 (y^2- 9y +22)dy = 115/6` .**

(y^2)-8y+13, y-x= 9, y = 2 and y = 7

y-[2,7] limites =2 to 7

(y^-9y+22)dy

y^3/3-9y^/2+22y

put the limits:upper limit - lower limit.

343/3-8/3-441/236/2+154-44

(686-16-1323+108+924-264)/6

115/6