1. ## Integration

Hi guys, im given this question:

With the given equation 2^t + t^2, find the area between 0 and 1 via:

i) finding certain limits.

ii) Fundamental theorem of calculus.

Can someone show me what to do for part i). I really dont know what to do.

For part ii, i was wondering if this is correct: Integrate 2^t and t^2. That becomes 2^t = 2^t / ln2 + t^3/3 and then put in the values of 0 and 1 respectively. Is it the right way?

Thanks guys

2. If you consider a definite integral as limit of a 'Riemann sum' is...

$\int_{0}^{1} f(t)\cdot dt = \lim_{n \rightarrow \infty} \frac{1}{n}\cdot \sum_{i=1}^{n} f(\frac{i}{n})$ (1)

In our case is $f(t)= 2^{t} + t^{2}$ so that...

$f(\frac{i}{n}) = 2^{\frac{i}{n}} + (\frac{i}{n})^{2}$ (2)

Now is...

$\sum_{i=1}^{n} 2^{\frac{i}{n}} = \frac{2-2^{\frac{1}{n}}}{2^{\frac{1}{n}}-1}$

$\sum_{i=1}^{n}(\frac{i}{n})^{2}= \frac{1}{n^{2}}\cdot \sum_{i=1}^{n} i^{2} = \frac {n\cdot (n+1)\cdot (2n+1)}{6\cdot n^{2}}$ (3)

... and...

$2^{\frac{1}{n}}-1= e^{\frac{\ln 2}{n}}-1 = \frac{\ln 2}{n} + \frac{\ln^{2} 2}{2\cdot n^{2}} + \dots$ (4)

... so that (1) becomes...

$\int_{0}^{1} (2^{t} + t^{2})\cdot dt = \lim_{n \rightarrow \infty} \frac{1}{n}\cdot (\frac {2-2^{\frac{1}{n}}}{2^{\frac{1}{n}}-1} + \frac{2\cdot n^{3} + 3\cdot n^{2} + n}{6\cdot n^{2}}) = \frac {1}{\ln 2} + \frac{1}{3}$ (5)

Kind regards

$\chi$ $\sigma$

3. Thanks a lot chisigma! So the Riemann sum is used to find certain limits?

As well, i was wondering in equation 3 how did u derive the numerator and denominator?

Thank you sooo much!

4. In the passage (3) you have to use the known result...

$\sum_{i=1}^{n} i^{2} = \frac{n\cdot (n+1)\cdot (2n+1)}{6}$

Kind regards

$\chi$ $\sigma$

5. Originally Posted by chisigma
In the passage (3) you have to use the known result...

$\sum_{i=1}^{n} i^{2} = \frac{n\cdot (n+1)\cdot (2n+1)}{6}$

Kind regards

$\chi$ $\sigma$
thanks again chisigma

Much appreciated