Question

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

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

Short answer

Hence proved

Step-by-step solution

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

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