Question

The area of the region bounded by the parabola y = x² + 1 and the straight line x + y = 3 is given by

Options

• \[\frac{45}{7}\]

• \[\frac{25}{4}\]

• \[\frac{\pi}{18}\]

• \[\frac{9}{2}\]

Answer 1

\[\frac{9}{2}\]

To find the point of intersection of the parabola y = x² + 1 and the line x + y = 3 substitute y = 3 − x in y = x² + 1
\[3 - x = x^2 + 1\]
\[ \Rightarrow x^2 + x - 2 = 0\]
\[ \Rightarrow \left( x - 1 \right)\left( x + 2 \right) = 0\]
\[ \Rightarrow x = 1\text{ or }x = - 2\]
\[ \therefore y = 2\text{ or }y = 5\]
So, we get the points of intersection A(−2, 5) and C(1, 2).
Therefore, the required area ABC,
\[A = \int_{- 2}^1 \left( y_1 - y_2 \right) dx ............\left(\text{Where, }y_1 = 3 - x\text{ and }y_2 = x^2 + 1 \right)\]
\[ = \int_{- 2}^1 \left[ \left( 3 - x \right) - \left( x^2 + 1 \right) \right] d x\]
\[ = \int_{- 2}^1 \left( 3 - x - x^2 - 1 \right) d x\]
\[ = \int_{- 2}^1 \left( 2 - x - x^2 \right) d x\]
\[ = \left[ 2x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{- 2}^1 \]
\[ = \left[ 2\left( 1 \right) - \frac{\left( 1 \right)^2}{2} - \frac{\left( 1 \right)^3}{3} \right] - \left[ 2\left( - 2 \right) - \frac{\left( - 2 \right)^2}{2} - \frac{\left( - 2 \right)^3}{3} \right]\]
\[ = \left[ 2 - \frac{1}{2} - \frac{1}{3} \right] - \left[ - 4 - 2 + \frac{8}{3} \right]\]
\[ = 2 - \frac{1}{2} - \frac{1}{3} + 4 + 2 - \frac{8}{3}\]
\[ = 8 - \frac{1}{2} - \frac{9}{3}\]
\[ = 5 - \frac{1}{2}\]
\[ = \frac{9}{2}\text{ square units }\]