The area bounded by the curve y = x^2 + 4x + | vedclass

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(B) The given curve is $y = x^2 + 4x + 5$. Completing the square,we get $y = (x+2)^2 + 1$.The minimum value of $y$ occurs at $x = -2$,where $y = 1$. T

Vedclass · Accepted Answer

$4\,\frac{2}{3}$

Vedclass · Answer

(B) The given curve is $y = x^2 + 4x + 5$. Completing the square,we get $y = (x+2)^2 + 1$.The minimum value of $y$ occurs at $x = -2$,where $y = 1$. This is the minimum ordinate.The curve intersects the $y$-axis at $x = 0$,where $y = 5$.The area bounded by the curve,the $y$-axis $(x=0)$,the $x$-axis $(y=0)$,and the minimum ordinate $(x=-2)$ is given by the integral:$A = \int_{-2}^{0} (x^2 + 4x + 5) \, dx$$= \left[ \frac{x^3}{3} + 2x^2 + 5x \right]_{-2}^{0}$$= (0) - \left( \frac{(-2)^3}{3} + 2(-2)^2 + 5(-2) \right)$$= - \left( -\frac{8}{3} + 8 - 10 \right)$$= - \left( -\frac{8}{3} - 2 \right) = \frac{8}{3} + 2 = \frac{8+6}{3} = \frac{14}{3} = 4\frac{2}{3}$ square units.

The area bounded by the curve $y = x^2 + 4x + 5$,the coordinate axes,and the minimum ordinate is:

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