Finding the limit of an integral with a Riemann sum Question
Let $f$ be a differentiable function on $[0, 1]$ such that $f(1) = 1$. If the derivative of $f$ is also continuous on $[0, 1]$, evaluate
$$\lim_{y\to\infty} \int^1_0yx^yf(x) \,dx$$
Attempt
I tried to approximate the integral as a Riemann sum
$$ \lim_{y\to\infty} \int^1_0yx^yf(x) \,dx \\
=\lim_{y\to\infty} \lim_{k\to\infty}\sum_{n=0}^k \left(\frac{1}{k}\right)y\left(\frac{n}{k}\right)^y f\left(\frac{n}{k}\right)
$$
I'm not sure how to continue, can I merge the limits by taking $y=k$?
Thanks in advance.
 A: First, notice that $yx^y\to 0$ on $[0,1[$ but not uniformly since at $x=1$ it goes to $+\infty$. This means that $x\mapsto yx^y$ gets concentrated around $x=1$ as $y\to +\infty$. So we can conjecture that : $$\int_0^1 yx^y f(x)\text dx \simeq \int_0^1 yx^y f(1)\text dx = f(1)$$ as $y\to+\infty$.
Now, to prove this, let $\epsilon >0$. Then, take $M = \sup_{[0,1]}|f|$, and $\delta>0$ such that :
$$\forall x\in[1-\delta,1], |f(x) - f(1)|<\epsilon$$
Then, we compute :
\begin{align}
\left|\int_0^1yx^yf(x)\text dx -f(1)\right| &\leq \left|\int_0^1yx^y(f(x)-f(1))\text dx \right| + \frac{1}{y+1}|f(1)| \\
&\leq \int_0^1yx^y|f(x)-f(1)|\text dx+ \frac{1}{y+1}|f(1)|\\
&\leq 2M\int_0^{1-\delta}yx^y\text{d}x + \int_{1-\delta}^1yx^y|f(x)-f(1)|\text dx + \frac{1}{y+1}|f(1)| \\
&\leq 2M \frac{y(1-\delta)^{y+1}}{y+1} + \epsilon \frac{y(1-(1-\delta)^{y+1})}{y+1} +\frac{1}{y+1}|f(1)| \\
\end{align}
Since the last equantitiy goes to $\epsilon$ as $y\to +\infty$, we see that for $y$ big enough, we have :
$$\left|\int_0^1yx^yf(x)\text dx -f(1)\right|\leq 2\epsilon$$
Therefore :
$$\lim_{y\to+\infty}\int_0^1yx^yf(x)\text dx =f(1)$$
