How Do You Calculate the Area Under a Complex Function?

[fatou123]

i am struggling with this question for quite few days now and i am nowhere near to an appropriate solution.

The previous ( related) question were regarding the the function
g(x)= 5+9x-2x^2 /(x^2 -6x+23)^3/2.

i have found the derivatives, the stationary point and also to classify them as local maximum and minimum which aare at x=-1/2 and x=5 .

this is the final question which i am struggling with

find the area bounded by the graph
y= (100(2x+1)(5 - x))/(x^2 -6x+23)^(5/2)
and below the x-axis. give your answer to four significant question.

I am only struggling to find the integral so i can calculate the area by myself.
so far i have found the integral of the numerator to be
100(2x+1)(5-x)
100(5+9x-2x^2)
500x+450x^2-(200/2)x^3

for the denominator i have found 2/7(x^2 -6x+23)^7/2 but i am pretty sure this is wrong.

thank you for the help.

 

[tiny-tim]

hi fatou123!

(try using the X² icon just above the Reply box )

… so far i have found the integral of the numerator to be
100(2x+1)(5-x)
100(5+9x-2x^2)
500x+450x^2-(200/2)x^3

i assume you mean (200/3)?

for the denominator i have found 2/7(x^2 -6x+23)^7/2 but i am pretty sure this is wrong.

sure is!

i] you cannot integrate a fraction by integrating the top and bottom separately

(it doesn't even begin to work …forget about it)​

ii] you'll need to simplify the bottom first …

either complete the square or susbtitute u = x - 3 ​

 

[fatou123]

thank you .

after completing the square for the bottom i end up with this.

y= 100(2x+1)(5-x)
((x-3)^2 -3^2) +23)^5/2)

do i have to do the integral by part ?

 

[tiny-tim]

do i have to do the integral by part ?

difficult to see how

try a trig substitution instead (always useful when there's a square inside a square-root)

 

[fatou123]

thank you again tiny tim i think i am starting to crack it with a little help well a big one to be honest !
so i gather that the trig substitution might be of the form

sqrt a^2 +b^2 x^2 x=a/b tan(theta)
so far i found the denominator to be to be sqrt((x-3)^2-14)^5) and i think by using the indice rule you get to sqrt((x-3)-14)^3 for the denominator. where u=(x-3) and du=3dx

 

[Ray Vickson]

i am struggling with this question for quite few days now and i am nowhere near to an appropriate solution.

The previous ( related) question were regarding the the function
g(x)= 5+9x-2x^2 /(x^2 -6x+23)^3/2.

i have found the derivatives, the stationary point and also to classify them as local maximum and minimum which aare at x=-1/2 and x=5 .

this is the final question which i am struggling with

find the area bounded by the graph
y= (100(2x+1)(5 - x))/(x^2 -6x+23)^(5/2)
and below the x-axis. give your answer to four significant question.

I am only struggling to find the integral so i can calculate the area by myself.
so far i have found the integral of the numerator to be
100(2x+1)(5-x)
100(5+9x-2x^2)
500x+450x^2-(200/2)x^3

for the denominator i have found 2/7(x^2 -6x+23)^7/2 but i am pretty sure this is wrong.

thank you for the help.

Do you mean g(x) = 5 + 9x - [2x^2/(x^2 - 6x + 23)^(3/2)]
which is actually what your formula says when parsed using standard rules, or do you mean g(x) = (5 + 9x - 2x^2)/(x^2 - 6x + 23)^(3/2)? If you mean the latter, you need to use brackets.

RGV

 

[fatou123]

soz i meant g(x) = (5 + 9x - 2x^2)/((x^2 - 6x + 23)^(3/2)).

i still haven't learn how to do integral by substitution with trigonometry yet.

 

[tiny-tim]

hifatou123!

(try using the X² icon just above the Reply box )

without trig substitution, you'll need to write it out as

A/√(x² - 6x + 23) + B(x-3)/√(x² - 6x + 23) + C/(√(x² - 6x + 23))³

and then integrate each part separately