Question

Make a sketch of the region {(x, y) : 0 ≤ y ≤ x² + 3; 0 ≤ y ≤ 2x + 3; 0 ≤ x ≤ 3} and find its area using integration.

Answer 1

\[R = \left\{ \left( x, y \right): 0 \leq y \leq x^2 + 3 , 0 \leq y \leq 2x + 3 , 0 \leq x \leq 3 \right\}\]
\[ R_1 = \left\{ \left( x, y \right) : 0 \leq y \leq x^2 + 3 \right\}\]
\[ R_2 = \left\{ \left( x, y \right) :  0 \leq y \leq 2x + 3 \right\}\]
\[ R_3 = \left\{ \left( x, y \right) : 0 \leq x \leq 3 \right\}\]
\[ \Rightarrow R = R_1 \cap R_2 \cap R_3 \]
y = x² + 3 is a upward opening  parabola with vertex A(0, 3).
Thus R₁ is the region above x-axis and  below the parabola
y = 2x + 3 is a straight line passing through A(0, 3) and cuts y-axis on (−3/2, 0).
Hence R₂ is the region above x-axis and below the line
x = 3  is a straight line parallel to y-axis, cutting x-axis at E(3, 0).
Hence R₃ is the region above x-axis  and to the left of the line x = 3.
Point of intersection of the parabola  and  y = 2x + 3 is given by solving the two equations
\[y = x^2 + 3\]
\[y = 2x + 3\]
\[ \Rightarrow x^2 + 3 = 2x + 3\]
\[ \Rightarrow x^2 - 2x = 0\]
\[ \Rightarrow x\left( x - 2 \right) = 0\]
\[ \Rightarrow x = 0 \text{ or }x = 2\]
\[ \Rightarrow y = 3\text{ or }y = 7\]
\[ \Rightarrow A\left( 0, 3 \right)\text{ and }B\left( 2, 7 \right) \text{ are points of intersection }\]
\[\text{ Also }, x = 3\text{ cuts the parabola at }C\left( 3, 12 \right)\]
\[\text{ and }x = 3\text{ cuts }y = 2x + 3\text{ at }D\left( 3, 9 \right)\]
 We require thearea of shaded region. 
\[\text{ Total shaded area }= \int_0^2 \left( x^2 + 3 \right)dx + \int_2^3 \left( 2x + 3 \right)dx\]
\[ = \left[ \frac{x^3}{3} + 3x \right]_0^2 + \left[ \frac{2 x^2}{2} + 3x \right]_2^3 \]
\[ = \left[ \frac{x^3}{3} + 3x \right]_0^2 + \left[ x^2 + 3x \right]_2^3 \]
\[ = \left[ \frac{8}{3} + 6 \right] + \left[ 9 + 9 - 4 - 6 \right]\]
\[ = \frac{8}{3} + 6 + 8\]
\[ = \frac{8 + 42}{3}\]
\[ = \frac{50}{3}\text{  sq . units }\]