Use python to solve the problem

6.18: The temperature of a light bulb Exel se bulb is a simple devi incande light it contains a fi ament lmally. heated by the flow of electricity until it becomes usually made Essentially all of the power hot enough to radi ie but some of the radiation is by a bulb not in the visible w as elec ngths, which toma t is useless for lighting purposes Let us define the efficienc to be the fraction of the r tam a in the ible band. It's a good a to assume that the radiation at temperature T Exercise 6.13 on page 267), meaning that the law gth obeys power radiated per unit wave. here A is the surface area of the filament, h is constant, c is the speed of ht, and ks is Boltzmann's constant. The visible wavel run from nm 750nm, so the total energy radiated the engths. in visible window I(A) da the dA. Dividing one expression by the other ght substituting for 101) from above, we get an expression for the efficiency n of the bulb thus 1) dA. 1) dA where the leading constants and the area A have canceled out. Making the substitution he/AkBT, this can also be written as x3/( 1) dx x3/(ex -1) dx where we have made use of the known exact value of the integral in the denominator. a) Write a Python function that takes a temperature T as its argument and calcu- lates the value of for that temperature from the formula above. The integral in the formula cannot be done analytically, but you can do it numerically using any method of your choice. (For instance, Gaussian quadrature with 100 sample Points works fine.) Use your function to make a graph of as a function of tem perature between 300 K and 10000K. You should see that there is an intermediate temperature where the efficiency is a maximum