Evaluate the following limit: $\lim\limits_{h\to 0} \frac {1}{h} \int_2^{2+h}\sin(x^2) \, dx $

Question

Evaluate the following limit: $$\lim_{h\to 0} \frac {1}{h} \int_2^{2+h} \sin(x^2) \, dx $$

Would this be correct?

\begin{align} & \lim_{h\to 0} \frac {1}{h} \int_2^{2+h} \sin(x^2) \, dx \\[10pt] = {} & \lim_{h\to0} \frac{ \int_0^{2+h} \sin(x^2) \, dx -\int_0^2 \sin(x^2) \, dx}{h}\\[10pt] = {} &\lim_{h\to0} \frac{f(2+h)-f(2)} h = f'(2) = {} \sin(2^2) = \sin 4 \end{align}

where

$$f(u) = \int_0^u \sin(x^2)\, dx.$$

Or must I define $f(u)$ as follows? $$f(u) = \int_2^u \sin(x^2)\, dx.$$

Answer 1

That would be $\sin 4$. You can use the fundamental theorem of calculus directly.

Answer 2

\begin{align} & \lim_{h\to 0} \frac {1}{h} \int_2^{2+h} \sin(x^2) \, dx \\[10pt] = {} & \lim_{h\to0} \frac{ \int_2^{2+h} \cdots -\int_2^2 \cdots }{h} \\[10pt] = {} &\lim_{h\to0} \frac{f(2+h)-f(2)} h = f'(2) \end{align} where $$ f(u) = \int_2^u \sin(x^2)\, dx. $$ So use the fundamental theorem of calculus to find $f'(2).$ (You get $\sin(2^2).$)

Answer 3

MVT for integrals :

$(1/h)\int_{2}^{2+h} \sin x^2 dx=$

$(1/h)\sin t^2 \int_{2}^{2+h}1 dx= \sin t^2,$

where $t \in [2,2+h]$.

Note: $ \lim_{ h \rightarrow 0} t = 2$.

Take the limit.