Question

Solve the given definite integral $I=\int_{\infty }^{R}{\dfrac{GMm}{{{x}^{2}}}}dx$.

Answer 1

Hint: We will apply the formula of integral here. The formula is given by $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c$ where c is any constant. But since, we are given the integral in the definite form. This is why we do not apply c here. Thus, we will apply $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}$ here. With the help of the integral formula, we will be able to solve the equation.

Complete step-by-step solution -
Now, we will consider the integral expression $I=\int_{\infty }^{R}{\dfrac{GMm}{{{x}^{2}}}}dx$...(i)
As we can clearly see that the dx is in terms of x. This means that except for x all will behave like constants. Thus we have G, M, and m as constants here and x is a variable. So, we will take out the terms G, M, and m from the integral part. Thus our new equation is given as $I=GMm\int_{\infty }^{R}{\dfrac{1}{{{x}^{2}}}}dx$. As $\dfrac{1}{{{x}^{2}}}$ can also be written as ${{x}^{-2}}$. Thus, we have $I=GMm\int_{\infty }^{R}{{{x}^{-2}}}dx$.
Now, we will apply the formula which is given by $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c$ where c is any constant. But since, we are given the integral in definite form. This is why we not apply c here. Thus, we will apply $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}$ here. With this formula we can convert $\int{{{x}^{-2}}}dx=\dfrac{{{x}^{-2+1}}}{-2+1}$. Now we will solve it further. Therefore we get $\int{{{x}^{-2}}}dx=\dfrac{{{x}^{-1}}}{-1}$ or $\int{{{x}^{-2}}}dx=-\dfrac{1}{x}$.
Now we will substitute this value in equation $I=GMm\int_{\infty }^{R}{{{x}^{-2}}}dx$. Thus, we get $I=GMm\left[ -\dfrac{1}{x} \right]_{\infty }^{R}$. As we have the values for x here that we will open by the formula $\left[ x \right]_{a}^{b}=x\left( b \right)-x\left( a \right)$. Thus, the equation $I=GMm\left[ -\dfrac{1}{x} \right]_{\infty }^{R}$ changes into $I=GMm\left( -\left( \dfrac{1}{R}-\dfrac{1}{\infty } \right) \right)$. As the value of $\dfrac{1}{\infty }=0$. Thus we get $I=GMm\left( -\left( \dfrac{1}{R} \right) \right)$ or $I=-\dfrac{GMm}{R}$.
Hence, the value of the integral $I=-\dfrac{GMm}{R}$.

Note: While coming across $\dfrac{1}{\infty }$ we will not stop just there. We will still solve by substituting $\dfrac{1}{\infty }=0$. Otherwise, we will not get the desired result. In case we have so many alphabets in an integral and we do not know what terms we are going to solve. So, for that, we will see the derivative sign. Here in this question, we have dx as a derivative sign with the integral. Therefore, as dx has x in it so, we will integrate in terms of x only. One can get confused while using the formula $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c$ with $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}+c$ where c is an arbitrary constant. So, learning and using the two formulas with practice will result in the right usage of these. Also, the difference between them will also be accessible.