The area enclosed by y=3x-5,y=0,x=3,and x=5 i | vedclass

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(D) The area $A$ enclosed by the curve $y=3x-5$,the $x$-axis $(y=0)$,and the lines $x=3$ and $x=5$ is given by the definite integral:$A = \int_{3}^{5}

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$14 	ext{ sq. units}$

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(D) The area $A$ enclosed by the curve $y=3x-5$,the $x$-axis $(y=0)$,and the lines $x=3$ and $x=5$ is given by the definite integral:$A = \int_{3}^{5} (3x-5) \, dx$Evaluating the integral:$A = \left[ \frac{3x^2}{2} - 5x ight]_{3}^{5}$Substitute the upper limit $x=5$:$\left( \frac{3(5)^2}{2} - 5(5) ight) = \left( \frac{75}{2} - 25 ight) = \frac{75-50}{2} = \frac{25}{2} = 12.5$Substitute the lower limit $x=3$:$\left( \frac{3(3)^2}{2} - 5(3) ight) = \left( \frac{27}{2} - 15 ight) = \frac{27-30}{2} = -\frac{3}{2} = -1.5$Calculate the final area:$A = 12.5 - (-1.5) = 12.5 + 1.5 = 14 	ext{ sq. units}$

The area enclosed by $y=3x-5$,$y=0$,$x=3$,and $x=5$ is

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