Area bounded by the curves \( y=3^x,y=x \) and \( x=1,x=2 \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \left(\displaystyle\frac{6}{\log_e 3}-\displaystyle\frac{3}{2}\right) \) Square units
On [1, 2], \( 3^x\gt x \), so the curve \( y=3^x \) is on top and the area is
\( \displaystyle \int_1^2(3^x-x)\,dx=\left[\frac{3^x}{\log_e 3}-\frac{x^2}{2}\right]_1^2 \)
\( \displaystyle =\frac{9-3}{\log_e 3}-\frac{4-1}{2} \)
\( \displaystyle =\frac{6}{\log_e 3}-\frac{3}{2} \)