Question:
Interpret
$\lim _{n\to \infty }\left(\sum _{k=1}^n\:6\sqrt{\frac{11}{n^2}-\frac{6k}{n^3}}\right)$
as a definite integral and evaluate it.
Attempt:
To interpret the given limit as a definite integral, we first recognize that it is a Riemann sum. The limit represents the sum of the function values, multiplied by the width of the intervals, as the number of intervals $(n)$ goes to infinity.
Given limit:
$$\lim_{n\to \infty} \left(\sum_{k=1}^n 6\sqrt{\frac{11}{n^2}-\frac{6k}{n^3}}\right)$$
Let's rewrite the expression inside the square root in terms of a single fraction:
$$6\sqrt{\frac{11n – 6k}{n^3}}$$
Now, let's identify the components of the Riemann sum:
$$\Delta x = \frac{1}{n}$$
$$x_k = \frac{k}{n}$$
As $n \to \infty$, the Riemann sum converges to a definite integral:
$$\int_{a}^{b} f(x) dx$$
Now, we need to find the function $f(x)$ and the limits of integration $[a,b]$. Since $x_k = k/n$, we can substitute this into the expression:
$$6\sqrt{\frac{11n – 6(k/n)}{n^3}}$$
Now, we can rewrite this expression in terms of $x$:
$$6\sqrt{\frac{11 – 6x}{n^2}}$$
And simplify:
$$6\sqrt{\frac{11 – 6x}{n^2}}$$
Now, we can identify the function $f(x) = 6\sqrt{11 – 6x}$. We also need to find the limits of integration, which are the values of $x$ corresponding to the first and last terms of the sum when $k = 1$ and $k = n$:
$$x_1 = \frac{1}{n} \to 0\ \text{as}\ n \to \infty\ (\text{lower limit},\ a)$$
$$x_n = \frac{n}{n} \to 1\ \text{as}\ n \to \infty\ (\text{upper limit},\ b)$$
Now we can write the definite integral:
$$\int_{0}^{1} 6\sqrt{11 – 6x} dx$$
To evaluate the integral, we can make a substitution:
Let $u = 11 – 6x$
$$\frac{du}{dx} = -6$$
$$dx = \frac{du}{-6}$$
The limits of integration will change accordingly:
$$x = 0 \Rightarrow u = 11$$
$$x = 1 \Rightarrow u = 5$$
Now we have:
$$-\int_{11}^{5} \sqrt{u} du$$
To evaluate this integral, we use the power rule for integration:
$$-\left[\frac{2}{3}u^{3/2}\right]_{11}^{5}$$
Evaluate the antiderivative at the limits of integration:
$$-\left(\frac{2}{3}(5^{3/2}) – \frac{2}{3}(11^{3/2})\right)$$
Simplify:
$$\frac{2}{3}(11^{3/2} – 5^{3/2})$$
So, the definite integral is:
$$\frac{2}{3}(11^{3/2} – 5^{3/2})$$
Is this correct ?
Best Answer
You can speed up the calculations by going directly to the formula $$\sum_{k=1}^n{6\over n}\sqrt{11-{6k\over n}}=\sum_{k=0}^{n-1}{6\over n}\sqrt{11-{6(n-k)\over n}}=\sum_{k=0}^{n-1}{6\over n}\sqrt{5+{6k\over n}}=\sum_{k=1}^{n}{6\over n}\sqrt{5+{6(k-1)\over n}}$$ which by taking $\Delta_n={6\over n} $ and the corresponding partition of the interval $[5,11]$ results in $$\int\limits_5^{11}\sqrt{x}\,dx$$ Mind that the last Riemann sum is of the type left hand side point of the partition interval.