The area of the region bounded by the parabol | vedclass

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(D) Given equation of the parabola is $y=x^{2}-4x+5$ $(i)$ and the equation of the line is $y=x+1$ (ii).To find the points of intersection, we equate

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$\frac{9}{2}$

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(D) Given equation of the parabola is $y=x^{2}-4x+5$ $(i)$ and the equation of the line is $y=x+1$ (ii).To find the points of intersection, we equate the two equations:$x^{2}-4x+5 = x+1$$x^{2}-5x+4 = 0$$(x-1)(x-4) = 0$Thus, the points of intersection are at $x=1$ and $x=4$.The required area is given by the integral of the upper curve minus the lower curve from $x=1$ to $x=4$:$	ext{Area} = \int_{1}^{4} \{(x+1) - (x^{2}-4x+5)\} dx$$	ext{Area} = \int_{1}^{4} (-x^{2}+5x-4) dx$$	ext{Area} = \left[ -\frac{x^{3}}{3} + \frac{5x^{2}}{2} - 4x ight]_{1}^{4}$Substituting the limits:$	ext{Area} = \left( -\frac{64}{3} + \frac{5(16)}{2} - 4(4) ight) - \left( -\frac{1}{3} + \frac{5(1)}{2} - 4(1) ight)$$	ext{Area} = \left( -\frac{64}{3} + 40 - 16 ight) - \left( -\frac{1}{3} + \frac{5}{2} - 4 ight)$$	ext{Area} = \left( 24 - \frac{64}{3} ight) - \left( \frac{5}{2} - \frac{13}{3} ight)$$	ext{Area} = \frac{8}{3} - \left( \frac{15-26}{6} ight) = \frac{8}{3} - \left( -\frac{11}{6} ight) = \frac{16+11}{6} = \frac{27}{6} = \frac{9}{2}$.

The area of the region bounded by the parabola $y=x^{2}-4x+5$ and the straight line $y=x+1$ is

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