Calculus 2. Calculate $\int_0^3 \left(x^2 +2\right) dx$ by the formal definition.

Question

Calculate by the formal definition of the definite Integral

$$ \int_0^3 \left(x^2 +2\right) dx.$$

I have

$\Delta x = \frac 3n,\ x_i^* = \frac {3i}n \\f(x_i) = (\frac {3i}n)^2 +2 $

Express $$ \sum_{k=1}^n f(x_i^*)\ \Delta x $$

in closed form, the answer will be in terms of $n$

The part i am stuck on is getting the constants out front of the $\sum$ sign.

I've plugged $x_i^* $ into $ f(x)$ and got $\frac {9i^2 + 2n}{n^2} $

but then i cant really figure out a way to isolate $i$ behind the $\sum$

Can you simplify (plug into $f$ and factor out anything that doesn't depend on $i$)? — Ian, Jan 16 '16 at 22:02

score 0 · Accepted Answer · edited Apr 13 '17 at 12:21

0

You have $$ \sum_{i=1}^n f(x_i^*)\ \Delta x=\frac {27}{n^3} \sum_{i=1}^n i^2+ \frac {6}{n}\sum_{i=1}^n 1 $$ then use $$ \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} $$ which is proved here to get, as $n \to \infty$, $$ \sum_{i=1}^n f(x_i^*)\ \Delta x \to 15 $$

edited Apr 13 '17 at 12:21

Community

1

answered Jan 16 '16 at 22:08

Olivier Oloa

120,989

Quick question, where did the $ \frac6n$ come from? – impact_sv1 Jan 16 '16 at 22:14
so the ending equation for Rn should be $\frac{27((n+1)(2n+1)))}{n^2} $ correct? – impact_sv1 Jan 16 '16 at 22:17
Rather:$\frac{27(n+1)(2n+1)}{6n^2}\sim \frac{27\times2}6=9.$ Then you add $6$. – Olivier Oloa Jan 16 '16 at 22:18
@impact_sv1 tell me if it is ok for you. Thanks. – Olivier Oloa Jan 16 '16 at 22:21
1

that worked ! Thanks dude! – impact_sv1 Jan 16 '16 at 22:24

Calculus 2. Calculate $\int_0^3 \left(x^2 +2\right) dx$ by the formal definition.

1 Answers1