Draw the rough sketch of the curve y = x^2 - 2x - 3 and x + y = 9 and using integration find the area between them.

The problem involves plotting the curves \( y = x^2 - 2x - 3 \) and \( x + y = 9 \), then determining the enclosed area between them using definite integration.

Step 1: Sketching the Curves

Quadratic Curve: \( y = x^2 - 2x - 3 \)

This equation represents a parabola. To sketch it, we find its roots by solving:

\( x^2 - 2x - 3 = 0 \)

Factorizing:

\( (x - 3)(x + 1) = 0 \)

So, the roots are \( x = 3 \) and \( x = -1 \), indicating x-intercepts.

The vertex of the parabola is found using the formula \( x = \frac{-b}{2a} \):

\( x = \frac{-(-2)}{2(1)} = 1 \)

Substituting \( x = 1 \) into the equation:

\( y = 1^2 - 2(1) - 3 = -4 \)

Thus, the vertex is \( (1, -4) \), and the parabola opens upwards.

Linear Equation: \( x + y = 9 \)

Rewriting in terms of \( y \):

\( y = 9 - x \)

This represents a straight line with a y-intercept at \( (0,9) \) and an x-intercept at \( (9,0) \).

Step 2: Finding Points of Intersection

Setting \( x^2 - 2x - 3 = 9 - x \):

\( x^2 - x - 12 = 0 \)

Factorizing:

\( (x - 4)(x + 3) = 0 \)

Thus, \( x = 4 \) and \( x = -3 \).

Substituting into \( y = 9 - x \), we get:

• For \( x = 4 \), \( y = 9 - 4 = 5 \) → (4,5)
• For \( x = -3 \), \( y = 9 - (-3) = 12 \) → (-3,12)

Step 3: Setting Up the Integral

The enclosed area is found by integrating the difference:

\( \int_{-3}^{4} [(9 - x) - (x^2 - 2x - 3)] dx \)

Simplifying the integrand:

\( 9 - x - x^2 + 2x + 3 = 12 + x - x^2 \)

Step 4: Evaluating the Integral

Computing:

\( \int_{-3}^{4} (12 + x - x^2) dx \)

Integrating each term:

\( \int 12 dx = 12x \), \( \int x dx = \frac{x^2}{2} \), \( \int x^2 dx = \frac{x^3}{3} \)

Applying limits:

\( \left[ 12x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-3}^{4} \)

Final Computation

Evaluating at \( x = 4 \):

\( 12(4) + \frac{4^2}{2} - \frac{4^3}{3} = 48 + 8 - \frac{64}{3} \)

\( = \frac{144}{3} + \frac{24}{3} - \frac{64}{3} = \frac{104}{3} \)

Evaluating at \( x = -3 \):

\( 12(-3) + \frac{(-3)^2}{2} - \frac{(-3)^3}{3} = -36 + \frac{9}{2} + \frac{27}{3} \)

\( = -36 + 4.5 + 9 = -22.5 \)

Computing the final result:

\( \frac{104}{3} - (-22.5) = \frac{104}{3} + \frac{45}{2} \)

Taking LCM of 6:

\( \frac{208}{6} + \frac{270}{6} = \frac{478}{6} = 79.67 \)

Conclusion

The area enclosed between the parabola and the straight line is approximately 79.67 square units.