Question

write the definite integral for the summation lim n to infinity of the summation k=1 to n (4+(3k)/n))^2(3/n)

Answer 1

The partial sum approximates \(\int_a^b f(x) dx\).

\(f\) has to be found inside the expression \(\sum_{k=1}^n (4+\frac{3k}n)^2 \frac3n\).

Something disappears: the last \(\frac{3}{n}\), it stands for \(x_{i+1}-x_i\) (see https://en.wikipedia.org/wiki/Riemann_integral#Riemann_sums)

Now, \((4+\frac{3k}n)^2\) is \(f(x)\) : \(x\) is approximated by \(4+\frac{3k}n=4+k\times \frac3n\).
=> \(f(x)=x^2\).
What is the smallest \(x\) in this sum? the largest?

Answer 2

look at \(x\approx 4+\frac{3k}n\) for smallest and \(k\) and largest \(k\) possible.

Answer 3

3k/n ?

Answer 4

informally: \(4+\frac{3k}{n}\) "becomes" \(x\).
Smallest value of \(4+\frac{3k}{n}\) is found when \(k=1\). -> \(4+\frac{3}{n}\). When \(n\) becomes large, this is "equal" to \(4\).

What is the largest possible value of \(4+\frac{3k}{n}\) ?

Answer 5

it is (look at the limits of the Rieman sum) when \(k=n\). What value of "\(x\)" does it give?

Answer 6

1?

Answer 7

or 4+3k/n?

Answer 8

yes (for the latter)..
"\(x\in [a,b]\)" is simulated by changing the value of \(k\) in \(4+\frac{3k}{n}\).
When \(k=n\), the value of \(4+\frac{3k}{n}\) is \(4+\frac{3n}{n} = 4+3=7\).

Answer 9

That means, the covered interval will be \([4, 7]\).

Answer 10

Thank you!