# calculus

use reimann sums and a limit to compute the exact area under the curve of y=2x^2+1 on the interval[1,3]... i am having trouble getting the right answer which is suppose to be 58/3. I would just like to see the set up of the sum so check if am doing it right. Thank you.

1. The Riemann sum is the basis of definite integration.

What we are trying to do is to divide the area to be integrated into n vertical pieces of equal width (along x). Call this width h.

We then express the area of each piece according to the definition of the function to be summed. The expression is then simplified and summed from 1 to n (using the right Riemann sum).

Finally, we take the limit for n-> ∞ which should be the exact area as required.

For
f(x) = 2x²+x
to be summed from 1 to 3.
Subdivide the area into n pieces, each of width h=(3-1)/n=2/n.

The x-value for the ith piece is then
1+ih = 1+2i/n.

The corresponding y-value is:
fi=f(1+ih)=f(1+2i/n)=2(1+2i/n)²+1+2i/n.

The width of the ith piece is h=2/n.

Thus the area of the ith piece is
h*fi = (2/n)[2(1+2i/n)²+(1+2i/n)]
Expand and simplify
h*fi=(6/n+20i/n²+8i²/n³)

The total area is
A=

n
Σh*fi
i=1

n
=Σ(6/n+20i/n²+8i²/n³)
i=1

= 6 + 20n(n+1)/(2n²)+8n(n+1)(2n+1)/(6n³)

As n->∞
A=6+10+16/6=18 1/3 = 58/3

Note:
Standard summations used:
n
Σ i = n(n+1)/2
i=1

n
Σ i² = n(n+1)(2n+1)/(6)
i=1

posted by MathMate

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