This would be the average value of R(x) between 10 m and 50 m :
[tex]\displaystyle \frac1{50\,\mathrm m - 10\,\mathrm m} \int_{10\,\rm m}^{50\,\rm m} kx^2 \, dx[/tex]
The average intensity of radiation between 10 meters and 50 meters from the source is 1033.33 k.
It is the reverse of differentiation.
The intensity of radiation at a distance x meters from a source is modeled by the function given by
R(x)=kx²
where k is a positive constant.
The average intensity of radiation between 10 meters and 50 meters from the source will be
[tex]\rm Average \ intensity = \dfrac{1}{50 - 10} \int_{10}^{50} \ kx^2 \ dx\\\\\\Average \ intensity = \dfrac{k}{50 - 10} [ \dfrac{x^3}{3} ]_{10}^{50}\\\\\\Average \ intensity = \dfrac{k}{40} [ \dfrac{50^3}{3} - \dfrac{10^3}{3} ]\\\\\\Average \ intensity =1033.33k[/tex]
More about the integration link is given below.
https://brainly.com/question/18651211
