`int_1^10(x - 4ln(x))dx` Express the integral as a limit of Riemann sums. Do not evaluate the limit.

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Chapter 5, 5.2 - Problem 30 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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lemjay | High School Teacher | (Level 3) Senior Educator

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`int_1^10 (x-4ln(x))dx`

To express this definite integral as limit of Riemann's Sum, apply the formula:

`int_a^b f(x) dx = lim_(n-> oo)sum_(i=1)^oo f(x_i)Delta x`

where

`Delta x = (b-a)/n`

`x_i = a + iDeltax`

For the given definite interval, the `Delta x` is:

`Delta x= (10-1)/n`

`Delta x=9/n`

And its `x_i` is:

`x_i= 1 + i *9/n`

`x_i=1+(9i)/n`

`x_i=(n+9i)/n`

Plug-in this x_i to the function.

The function in the given integral is:

`f(x) = x-4lnx`

So,  `f(x_i)` is:

`f(x_i) = (n+9i)/n-4ln((n+9i)/n)`

So the given definite integral

`int_1^10 (x-4lnx) dx`

becomes:

= `lim_(n->oo) sum_(i=1)^n ((n+9i)/n -4ln((n+9i)/n)) * 9/n`  

`=lim_(n->oo) sum_(i=1)^n (9(n+9i))/n^2-36/nln((n+9i)/n)`

 

Therefore, `int_1^10 (x-4lnx) dx=lim_(n-gtoo) sum_(i=1)^n (9(n+9i))/n^2-36/nln((n+9i)/n)` .

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