Multiple-choice question asking the student to identify which integral expression correctly represents the area of a described region, without requiring full evaluation.
The area of the region bounded by the straight lines $x = \frac{1}{2}$ and $x = 2$, and the curves given by the equations $y = \log_e x$ and $y = 2^x$ is (A) $\frac{1}{\log_e 2}(4 + \sqrt{2}) - \frac{5}{2}\log_e 2 + \frac{3}{2}$ (B) $\frac{1}{\log_e 2}(4 - \sqrt{2}) - \frac{5}{2}\log_e 2$ (C) $\frac{1}{\log_e 2}(4 - \sqrt{2}) - \frac{5}{2}\log_e 2 + \frac{3}{2}$ (D) none of the above
The area of the region bounded by the straight lines $x = \frac{1}{2}$ and $x = 2$, and the curves given by the equations $y = \log_e x$ and $y = 2^x$ is (A) $\frac{1}{\log_e 2}(4 + \sqrt{2}) - \frac{5}{2} \log_e 2 + \frac{3}{2}$ (B) $\frac{1}{\log_e 2}(4 - \sqrt{2}) - \frac{5}{2} \log_e 2$ (C) $\frac{1}{\log_e 2}(4 - \sqrt{2}) - \frac{5}{2} \log_e 2 + \frac{3}{2}$ (D) none of the above
Let $f$ and $g$ be two real-valued continuous functions defined on the closed interval $[a, b]$, such that $f(a) < g(a)$ and $f(b) > g(b)$. Then the area enclosed between the graphs of the two functions and the lines $x = a$ and $x = b$ is always given by (A) $\int_a^b |f(x) - g(x)| \, dx$ (B) $\left|\int_a^b (f(x) - g(x)) \, dx\right|$ (C) $\left|\int_a^b (|f(x)| - |g(x)|) \, dx\right|$ (D) $\int_a^b ||f(x)| - |g(x)|| \, dx$.
The area of the region between the curves $y = \sqrt{\frac{1+\sin x}{\cos x}}$ and $y = \sqrt{\frac{1-\sin x}{\cos x}}$ bounded by the lines $x = 0$ and $x = \frac{\pi}{4}$ is (A) $\int_0^{\sqrt{2}-1} \frac{t}{(1+t^2)\sqrt{1-t^2}}\,dt$ (B) $\int_0^{\sqrt{2}-1} \frac{4t}{(1+t^2)\sqrt{1-t^2}}\,dt$ (C) $\int_0^{\sqrt{2}+1} \frac{4t}{(1+t^2)\sqrt{1-t^2}}\,dt$ (D) $\int_0^{\sqrt{2}+1} \frac{t}{(1+t^2)\sqrt{1-t^2}}\,dt$
Consider the functions defined implicitly by the equation $y ^ { 3 } - 3 y + x = 0$ on various intervals in the real line. If $x \in ( - \infty , - 2 ) \cup ( 2 , \infty )$, the equation implicitly defines a unique real valued differentiable function $y = f ( x )$. If $x \in ( - 2,2 )$, the equation implicitly defines a unique real valued differentiable function $y = g ( x )$ satisfying $g ( 0 ) = 0$. The area of the region bounded by the curve $y = f ( x )$, the $x$-axis, and the lines $x = a$ and $x = b$, where $- \infty < a < b < - 2$, is (A) $\int _ { a } ^ { b } \frac { x } { 3 \left( ( f ( x ) ) ^ { 2 } - 1 \right) } d x + b f ( b ) - a f ( a )$ (B) $\quad - \int _ { a } ^ { b } \frac { x } { 3 \left( ( f ( x ) ) ^ { 2 } - 1 \right) } d x + b f ( b ) - a f ( a )$ (C) $\int _ { a } ^ { b } \frac { x } { 3 \left( ( f ( x ) ) ^ { 2 } - 1 \right) } d x - b f ( b ) + a f ( a )$ (D) $- \int _ { a } ^ { b } \frac { x } { 3 \left( ( f ( x ) ) ^ { 2 } - 1 \right) } d x - b f ( b ) + a f ( a )$
Area of the region bounded by the curve $y = e ^ { x }$ and lines $x = 0$ and $y = e$ is (A) $e - 1$ (B) $\int _ { 1 } ^ { e } \ln ( e + 1 - y ) d y$ (C) $e - \int _ { 0 } ^ { 1 } e ^ { x } d x$ (D) $\int _ { 1 } ^ { e } \ln y \, d y$
The function $f$ is one-to-one, and the shaded region between the lines $y = x$ and $x = 1$ and the curve $y = f ( x )$ in the first quadrant is given below. Which of the following is the expression of the area of the shaded region in terms of $\mathbf { f } ^ { - \mathbf { 1 } } ( \mathbf { x } )$? A) $\int _ { 0 } ^ { 2 } f ^ { - 1 } ( x ) d x$ B) $\int _ { 0 } ^ { 2 } \left( 2 - f ^ { - 1 } ( x ) \right) d x$ C) $\int _ { 0 } ^ { 1 } \left( x - f ^ { - 1 } ( x ) \right) d x$ D) $\int _ { 0 } ^ { 1 } \left( 2 - f ^ { - 1 } ( x ) \right) d x + \int _ { 1 } ^ { 2 } f ^ { - 1 } ( x ) d x$ E) $\int _ { 0 } ^ { 1 } \left( x - f ^ { - 1 } ( x ) \right) d x + \int _ { 1 } ^ { 2 } \left( 1 - f ^ { - 1 } ( x ) \right) d x$