Q. 73 - Calculus

Question

Prove (a) $\int e^{kx}dx=\frac{1}{k}{e}^{kx}+C \mathrm{using} \mathrm{differentiation} \mathrm{formulas}$

(b) $\int b^{x}dx=\frac{1}{\ln b}{b}^x+C \mathrm{by} \mathrm{differentiation} \mathrm{formulas}$

Step-by-Step Solution

Answer

Hence proved

Step 1. To prove

(a) $\int e^{kx}dx=\frac{1}{k}{e}^{kx}+C$

(b) $\int b^{x}dx=\frac{1}{\ln b}{b}^x+C$

Part(a) Step 2. Proving

$$\begin{gathered} \int e^{kx}dx=\frac{1}{k}{e}^{kx}+C \\ Now, \\ \frac{d}{dx}\left(\frac{1}{k}\right)e^{kx} \\ =\left(\frac{1}{k}\right)k{e}^{kx} \\ =e^{kx} \\ so, \frac{1}{k}{e}^{kx} \mathrm{is} \mathrm{an} \mathrm{antiderivative} \mathrm{of} {\mathrm{e}}^{\mathrm{kx}} \\ \mathrm{Hence} \mathrm{proved} \end{gathered}$$

Part( b) Step 3. Proving

$$\begin{gathered} \int b^{x}dx=\frac{1}{\ln b}{b}^x+C \\ \frac{d}{dx}\left(\frac{1}{\ln b}{b}^x\right) \\ =\left(\frac{1}{\ln b}{b}^x\right)\ln b \\ =b^x \\ \frac{1}{\ln b}{b}^x \mathrm{is} \mathrm{an} \mathrm{antiderivative} \mathrm{of} b^x. \\ \mathrm{Hence} \mathrm{proved} \end{gathered}$$