Question

Find area enclosed by $y=x^2+x+1,y=2x+3$.

Answer 1

9/2

Answer 2

To find the area enclosed by the two curves $y=x^2+x+1$ and $y=2x+3$, we follow these steps:

1. Find the points of intersection: Set the two equations equal to each other to find the x-coordinates where they intersect: $x^2+x+1 = 2x+3$ $x^2+x-2x+1-3 = 0$ $x^2-x-2 = 0$ Factor the quadratic equation: $(x-2)(x+1) = 0$ The intersection points occur at $x=-1$ and $x=2$. These will be our limits of integration.

2. Determine which curve is above the other: To determine which function is greater in the interval $[-1, 2]$, pick a test point, for example, $x=0$. For $y=x^2+x+1$, at $x=0$, $y=0^2+0+1=1$. For $y=2x+3$, at $x=0$, $y=2(0)+3=3$. Since $3 > 1$, the line $y=2x+3$ is above the parabola $y=x^2+x+1$ in the interval $[-1, 2]$.

3. Set up the definite integral: The area $A$ enclosed by the curves is given by the integral of the upper curve minus the lower curve, from the lower x-limit to the upper x-limit: $A = \int_{-1}^{2} ((2x+3) - (x^2+x+1)) dx$ $A = \int_{-1}^{2} (-x^2+x+2) dx$

4. Evaluate the integral: Integrate term by term: $A = \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^{2}$ Now, substitute the upper limit and subtract the result of substituting the lower limit: $A = \left( -\frac{(2)^3}{3} + \frac{(2)^2}{2} + 2(2) \right) - \left( -\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2(-1) \right)$ $A = \left( -\frac{8}{3} + \frac{4}{2} + 4 \right) - \left( -\frac{-1}{3} + \frac{1}{2} - 2 \right)$ $A = \left( -\frac{8}{3} + 2 + 4 \right) - \left( \frac{1}{3} + \frac{1}{2} - 2 \right)$ $A = \left( -\frac{8}{3} + 6 \right) - \left( \frac{2+3-12}{6} \right)$ $A = \left( \frac{-8+18}{3} \right) - \left( \frac{-7}{6} \right)$ $A = \frac{10}{3} + \frac{7}{6}$ To add the fractions, find a common denominator (6): $A = \frac{20}{6} + \frac{7}{6}$ $A = \frac{27}{6}$ Simplify the fraction: $A = \frac{9}{2}$

The area enclosed by the curves is $\frac{9}{2}$ square units.