Q. 72 - Calculus

Question

Prove $\int e^{x}dx=e^x+C by u\sin g \int e^{kx}dx=\frac{1}{k}{e}^{kx}+C$

and $\int e^{x}dx=e^x+C by u\sin g \int b^{x}dx=\frac{1}{\ln b}{b}^x+C$

Step-by-Step Solution

Answer

Hence Proved.

Step 1. To prove

(a) $\int e^{x}dx=e^x+C by u\sin g \int e^{kx}dx=\frac{1}{k}{e}^{kx}+C$

(b) $\int e^{x}dx=e^x+C by u\sin g \int b^{x}dx=\frac{1}{\ln b}{b}^x+C$

Part (a) Step 2. Proving

$$\begin{gathered} \int e^{x}dx=e^x+C \mathrm{by} \mathrm{using} \int e^{kx}dx=\frac{1}{k}{e}^{kx}+C \\ =\int e^{x}dx=\frac{1}{k}{e}^{kx}+C \\ =\int e^{x\left(1\right)}dx=\frac{1}{k}{e}^{x\left(1\right)}+C \\ =\int e^{x}dx=e^x+C \\ \mathrm{Hence} \mathrm{proved} \end{gathered}$$

Part( b) Step 3. Proving

$$\begin{gathered} \int e^{x}dx=e^x+C by u\sin g \int b^{x}dx=\frac{1}{\ln b}{b}^x+C \\ = \int b^{x}dx=\frac{1}{\ln b}{b}^x+C \\ = \int e^{x}dx=\frac{1}{\ln e}{e}^x+C \\ = \int e^{x}dx=\frac{1}{1}{e}^x+C \\ =\int e^{x}dx=e^x+C \\ \mathrm{Hence} \mathrm{proved} \end{gathered}$$