MHT CET · Maths · Area Under Curves
The area of the region bounded by the curve \(y=\log x, x\) -axis and the lines
\(x=1, x=e\) is
- A \(\frac{1}{e}\) Sq. Units
- B 1 Sq. Units
- C 4 sq. Units
- D \(\frac{1}{2}\) Sq. Units
Answer & Solution
Correct Answer
(B) 1 Sq. Units
Step-by-step Solution
Detailed explanation
Required area is shaded.
\(
\begin{aligned}
\therefore \mathrm{A} &=\int_{1}^{\mathrm{e}} \log \mathrm{x} \mathrm{dx}=\int_{1}^{\mathrm{e}}(\log \mathrm{x})(1) \mathrm{dx} \\
&=[\mathrm{x} \log (\mathrm{x})]_{1}^{\mathrm{e}}-\int_{1}^{\mathrm{e}} \frac{1}{\mathrm{x}} \times \mathrm{x} \mathrm{dx} \\
&=[\mathrm{e}(\log \mathrm{e})-0]-[\mathrm{x}]_{1}^{\mathrm{e}}=\mathrm{e}-(\mathrm{e}-1)=1
\end{aligned}
\)

\(
\begin{aligned}
\therefore \mathrm{A} &=\int_{1}^{\mathrm{e}} \log \mathrm{x} \mathrm{dx}=\int_{1}^{\mathrm{e}}(\log \mathrm{x})(1) \mathrm{dx} \\
&=[\mathrm{x} \log (\mathrm{x})]_{1}^{\mathrm{e}}-\int_{1}^{\mathrm{e}} \frac{1}{\mathrm{x}} \times \mathrm{x} \mathrm{dx} \\
&=[\mathrm{e}(\log \mathrm{e})-0]-[\mathrm{x}]_{1}^{\mathrm{e}}=\mathrm{e}-(\mathrm{e}-1)=1
\end{aligned}
\)

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