Find the area of the region R that lies under the given curbe y=f(x) over the indicated interval a < x < b. Under y= 11/squareroot(25-2x), over -48 < x < 0  area= ? Note: < has an...

Find the area of the region R that lies under the given curbe y=f(x) over the indicated interval a < x < b. Under y= 11/squareroot(25-2x),
over -48 < x < 0 

area= ?

Note: < has an equal sign under

Asked on by mavig

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the definite integral of the function `y = 11/sqrt(25-2x)` , such that:

`int_(-48)^0 11/sqrt(25-2x) dx`

You need to come up with the following substitution, such that:

`25 - 2x = u => -2dx = du => dx = -(du)/2`

You need to change the limits of integration, such that:

`25 - 2*(-48) = u_1 => u_1 = 121`

`25 - 2*0 = u_2 => u_2 = 25`

Changing the variable, yields:

`int_121^25 11/(sqrt u)*(-(du)/2)`

You need to use the following property of integrals, such that:

`int_a^b f(x)dx = - int_b^a f(x) dx`

`int_121^25 11/(sqrt u)*(-(du)/2) = int_25^121 11/(sqrt u)*((du)/2)`

`int_25^121 11/(sqrt u)*((du)/2) = (11/2)*int_25^121 1/(sqrt u)*(du)`

`(11/2)*int_25^121 1/(sqrt u)*(du) = (11/2)*u^(-1/2+1)/(-1/2+1)|_25^121`

`(11/2)*int_25^121 1/(sqrt u)*(du) = 11(sqrt u)|_25^121`

`(11/2)*int_25^121 1/(sqrt u)*(du) = 11(sqrt121 - sqrt25)`

`(11/2)*int_25^121 1/(sqrt u)*(du) = 11(11 - 5)`

`(11/2)*int_25^121 1/(sqrt u)*(du) = 66`

Hence, evaluating the area of the region under the given curve, over the given interval, yields `int_(-48)^0 11/sqrt(25-2x) dx = 66.`

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