Given f(x)=x^2+3, find the exact area A of the region under y=f(x) on interval [1,3] by computing the sum,then taking the limit as n goes to infinity. Thanks! 

Expert Answers

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To find the area under this curve in the given interval, you have to find the the limit of sum of areas of n rectangles as n approaches to infinity.

The width of the rectangle would be `Deltax` and the height of the rectangle is `f(x_i^*)` .

The range is from 1 to 3.

Let's find `Deltax` ,

`Deltax = (3-1)/n = 2/n`

Therefore, `x_i = 1+(2i0/n`

`f(x_i^*) = (x_i^*)^2+3`

`f(x_i^*) = (1+(2i)/n)^2+3`

`f(x_i^*) = (1+(4i)/n+(4i^2)/n^2)+3`

`f(x_i^*) = 4+(4i)/n+(4i^2)/n^2`


The area of a small rectangle is

`A = f(x_i^*) Deltax`

`A = (4+(4i)/n+(4i^2)/n^2)*(2/n)`

`A = 8/n+(8i)/n^2+(8i^2)/n^3`

THe sum of the ares of small rectangles is, `sumA`

`sum_(i=1)^nA = sum_(i=1)^n(8/n+(8i)/n^2+(8i^2)/n^3)`

`sum_(i=1)^nA = 8/nsum_(i=1)^n1+8/n^2sum_(i=1)^ni+8/n^3sum_(i=1)^ni^2`

Using summation identities,

`sum_(i=1)^nA = 8/n(n)+8/n^2(n^2/2+n/2)+8/n^3(n^3/3+n^2/2+n/6)`

`sum_(i=1)^nA = 8+4(1+1/n)+8(1/3+1/(2n)+1/(6n^2))`

`sum_(i=1)^nA = 44/3+8/n+4/(3n^2)`


The area under the curve is the limit of this,


`Area = lim_(n-gtoo)(44/3+8/n+4/(3n^2))`

Area = 44/3


Therefore the value of the integral is 44/3.


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