Question

$\lim_{n \to \infty} n^2 \int_{-1/n}^{1/n} (2010\sin x + 2012 \cos x) |x| dx$

Answer 1

2012

Answer 2

Let the integral be denoted by $I_n$: $I_n = \int_{-1/n}^{1/n} (2010\sin x + 2012 \cos x) |x| dx$

We can split the integral based on the definition of $|x|$: $I_n = \int_{-1/n}^{0} (2010\sin x + 2012 \cos x) (-x) dx + \int_{0}^{1/n} (2010\sin x + 2012 \cos x) (x) dx$

In the first integral, let $u = -x$. Then $du = -dx$. When $x = -1/n$, $u = 1/n$. When $x = 0$, $u = 0$. The first integral becomes: $\int_{1/n}^{0} (2010\sin(-u) + 2012 \cos(-u)) (u) (-du)$ $= \int_{1/n}^{0} (-2010\sin u + 2012 \cos u) (u) (-du)$ $= \int_{0}^{1/n} (-2010\sin u + 2012 \cos u) u du$

Replacing the variable $u$ with $x$: $= \int_{0}^{1/n} (-2010x\sin x + 2012x\cos x) dx$

Now, substitute this back into the expression for $I_n$: $I_n = \int_{0}^{1/n} (-2010x\sin x + 2012x\cos x) dx + \int_{0}^{1/n} (2010x\sin x + 2012x\cos x) dx$ $I_n = \int_{0}^{1/n} [(-2010x\sin x + 2012x\cos x) + (2010x\sin x + 2012x\cos x)] dx$ $I_n = \int_{0}^{1/n} (4024x\cos x) dx$ $I_n = 4024 \int_{0}^{1/n} x\cos x dx$

To evaluate the integral $\int x\cos x dx$, we use integration by parts ($\int u dv = uv - \int v du$): Let $u = x$ and $dv = \cos x dx$. Then $du = dx$ and $v = \sin x$. $\int x\cos x dx = x\sin x - \int \sin x dx = x\sin x - (-\cos x) = x\sin x + \cos x$.

Now, evaluate the definite integral: $I_n = 4024 [x\sin x + \cos x]_{0}^{1/n}$ $I_n = 4024 \left[\left(\frac{1}{n}\sin\left(\frac{1}{n}\right) + \cos\left(\frac{1}{n}\right)\right) - (0\sin(0) + \cos(0))\right]$ $I_n = 4024 \left(\frac{1}{n}\sin\left(\frac{1}{n}\right) + \cos\left(\frac{1}{n}\right) - 1\right)$

We need to find the limit of $n^2 I_n$ as $n \to \infty$: $\lim_{n \to \infty} n^2 I_n = \lim_{n \to \infty} n^2 \cdot 4024 \left(\frac{1}{n}\sin\left(\frac{1}{n}\right) + \cos\left(\frac{1}{n}\right) - 1\right)$ Let $h = 1/n$. As $n \to \infty$, $h \to 0$. The expression becomes: $4024 \lim_{h \to 0} \frac{1}{h^2} \left(h\sin h + \cos h - 1\right)$ $= 4024 \lim_{h \to 0} \frac{h\sin h + \cos h - 1}{h^2}$

This is an indeterminate form of type $\frac{0}{0}$. We can use Taylor series expansions for $\sin h$ and $\cos h$ around $h=0$: $\sin h = h - \frac{h^3}{6} + O(h^5)$ $\cos h = 1 - \frac{h^2}{2} + \frac{h^4}{24} + O(h^6)$

Substitute these into the numerator: $h\sin h + \cos h - 1 = h\left(h - \frac{h^3}{6} + ...\right) + \left(1 - \frac{h^2}{2} + \frac{h^4}{24} + ...\right) - 1$ $= h^2 - \frac{h^4}{6} + ... + 1 - \frac{h^2}{2} + \frac{h^4}{24} + ... - 1$ $= \left(h^2 - \frac{h^2}{2}\right) + \left(-\frac{h^4}{6} + \frac{h^4}{24}\right) + ...$ $= \frac{h^2}{2} - \frac{3h^4}{24} + ...$ $= \frac{h^2}{2} - \frac{h^4}{8} + ...$

Now, substitute this back into the limit: $4024 \lim_{h \to 0} \frac{\frac{h^2}{2} - \frac{h^4}{8} + ...}{h^2}$ $= 4024 \lim_{h \to 0} \left(\frac{1}{2} - \frac{h^2}{8} + ...\right)$ $= 4024 \left(\frac{1}{2}\right)$ $= 2012$

Alternatively, using L'Hopital's Rule on $\lim_{h \to 0} \frac{h\sin h + \cos h - 1}{h^2}$: Applying L'Hopital's rule once: $\lim_{h \to 0} \frac{\frac{d}{dh}(h\sin h + \cos h - 1)}{\frac{d}{dh}(h^2)} = \lim_{h \to 0} \frac{\sin h + h\cos h - \sin h}{2h}$ $= \lim_{h \to 0} \frac{h\cos h}{2h} = \lim_{h \to 0} \frac{\cos h}{2} = \frac{1}{2}$

The total limit is $4024 \times \frac{1}{2} = 2012$. The final answer is $\boxed{2012}$.