The area bounded by the curve y=x^2+3,y=x,x=3 | vedclass

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(C) The area $A$ bounded by the curves $y=f(x)$ and $y=g(x)$ between $x=a$ and $x=b$ is given by $A = \int_{a}^{b} |f(x) - g(x)| \, dx$. Here,the curv

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$\frac{27}{2}$ sq. units

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(C) The area $A$ bounded by the curves $y=f(x)$ and $y=g(x)$ between $x=a$ and $x=b$ is given by $A = \int_{a}^{b} |f(x) - g(x)| \, dx$. Here,the curves are $y = x^2 + 3$ and $y = x$. The region is bounded by the $y$-axis $(x=0)$ and the line $x=3$. Since $x^2 + 3 > x$ for all $x \in [0, 3]$,the area is: $A = \int_{0}^{3} (x^2 + 3 - x) \, dx$ $A = [\frac{x^3}{3} + 3x - \frac{x^2}{2}]_{0}^{3}$ $A = (\frac{27}{3} + 3(3) - \frac{9}{2}) - (0)$ $A = 9 + 9 - 4.5 = 18 - 4.5 = 13.5$ $A = \frac{27}{2}$ sq. units.

The area bounded by the curve $y=x^2+3$,$y=x$,$x=3$ and the $y$-axis is:

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