Definite integrals (Riemann sums)

Question

can anyone please help with th is

Saw the problem from schaums outline advanced calculus

Converting that to the limit of a sum I got:

The sum I got

My question is how did that translate to The sum I got

Answer 1

The definition of the integral as a Riemann sum is $$\int_a^bf(x)\,dx = \lim_{n \rightarrow \infty} \text{RS}(f,a,b,n) =\lim_{n \rightarrow \infty} \sum_{i = 0}^{n - 1} f \left( a + i \cdot \frac{b-a}{n} \right) \frac{b - a}{n}$$ It seems like in the sum you got the upper bound is $n$ instead of $n-1$. Taking the upper bound to be $n-1$, we get the sum $$\frac{1}{n} \sum_{k=1}^{n-1} \sin\left( \frac{kt}{n} \right)$$ Rewriting this we get $$\frac{1}{n} \sum_{k=1}^{n-1} \sin\left( \frac{kt}{n} \right) = \frac{1-0}{n} \sum_{k=1}^{n-1} \sin \left( 0 + k\frac{t - 0}{n} \right)$$ Since if $k=0$, $\displaystyle{\sin \left( 0 + k\frac{t-0}{n} \right)} = 0$. We can just set the lower bound to $k=0$. Continuing, we have $$\frac{1-0}{n} \sum_{k=1}^{n-1} \sin \left( 0 + k\frac{t - 0}{n} \right) = \frac{1-0}{n} \sum_{k=0}^{n-1} \sin \left( 0 + k\frac{t - 0}{n} \right)$$ Now, we manipulate the fraction at the start to get $$\frac{1-0}{n} \sum_{k=0}^{n-1} \sin \left( 0 + k\frac{t - 0}{n} \right) = \frac{1}{t} \frac{t-0}{n} \sum_{k=0}^{n-1} \sin \left( 0 + k\frac{t - 0}{n} \right)$$ Next, we can rewrite this sum as an integral from $0$ to $t$: $$\frac{1}{t} \frac{t-0}{n} \sum_{k=0}^{n-1} \sin \left( 0 + k\frac{t - 0}{n} \right) = \frac{1}{t} \int_0^t \sin(x) \, dx$$ Finally, we can see that if we want the Riemann sum of the integral $$\int_0^t \sin(x) \, dx$$ We need to multiply both sides of our equation by $t$: $$\frac{1}{t} \frac{t-0}{n} \sum_{k=0}^{n-1} \sin \left( 0 + k\frac{t - 0}{n} \right) = \frac{1}{t} \int_0^t \sin(x) \, dx \Rightarrow$$ $$\int_0^t \sin(x) \, dx = \frac{t-0}{n} \sum_{k=0}^{n-1} \sin \left( 0 + k\frac{t - 0}{n} \right)$$ Which after simplification equals $$\frac{t}{n}\sum_{k=1}^{n-1}\sin \left( \frac{kt}{n} \right)$$