Approximation of an integral (another)

68 Views Asked by At

I am reading a book and it makes me nuts. The authors say that there is a "well-known approximation", but I don't know it. I'll cite their passage:

"Next, using the well-known approximation of the integral

$\int_{a(t)} ^{a(t)+\delta a(t)} \beta(\tau)m(\tau)d\tau=\beta(a(t))m(a(t))\delta a(t)+o(\delta a(t))$

..."

Source: N. Hritonenko and Yu. Yatsenko, Mathematical Modeling in Economics, Ecology and the Environment, 2nd Edition, Springer, New York/Berlin, 2013, 296p· The passage is on page 127.

Would someone be so kind and answer me the following questions: - Which approximation rule is this? - Is there a nice textbook, in which it is simple explained? Thx Kerim

2

There are 2 best solutions below

2
On

Assuming that $f(x)$ is a reasonable function, we have $\;\int_a^b f(x)dx \approx f(a)(b-a)\;$ as a reasonable approximation. Another way of writing the approximation is $\int_a^b f(x)dx = f(a)(b-a) + o(b-a)$ using little o notation. Notice that here $f(a)$ can be replaced by $f(c)$ for any $c\in[a,b].$

0
On

"$\delta$" is usually used to indicate a small, often infinitesimal, change in something, so $\delta(a(\tau))$ is a small change in $a(\tau)$. Thus, $$ \int_{a(t)}^{a(t)+\delta(a(t))}\beta(\tau)m(\tau)\,\mathrm{d}\tau\tag1 $$ is the integral of $\beta(\tau)m(\tau)$ over the infinitesimal interval $$ \left[a(t),a(t)+\delta(a(t))\right]\tag2 $$ Assuming $\beta(\tau)m(\tau)$ is continuous, we have $$ \beta(\tau)m(\tau)=\beta(a(t))m(a(t))+o(1)\tag3 $$ over the interval in $(2)$, where $o(1)\to0$ as $\delta(a(t))\to0$. Integrating the approximation in $(3)$, we get $$ \int_{a(t)}^{a(t)+\delta(a(t))}\beta(\tau)m(\tau)\,\mathrm{d}\tau=\beta(a(t))m(a(t))\delta(a(t))+o(\delta(a(t)))\tag4 $$