Calculus

Application of Integrals

Medium

core

The area bounded by $2x+y=8, y=a$ and $y=a+2$ with y-axis is 5 sq.units where $a>0$ , then the value of $a$ is:

✅ Correct Option: 2

From the line, $x = \dfrac{8-y}{2}$ , so the area is $\int_a^{a+2}\dfrac{8-y}{2}\,dy$ .

Evaluating gives $\dfrac{1}{2}\left[16 - \dfrac{(a+2)^2-a^2}{2}\right] = \dfrac{1}{2}[16-2a-2] = 7-a$ .

Setting $7-a=5$ gives $a=2$ .

2022: 7 Aug Shift 2

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The area enclosed by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is given by :

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The area bounded by 2x+y=8,y=a2x+y=8, y=a2x+y=8,y=a and y=a+2y=a+2y=a+2 with y-axis is 5 sq.units where a>0a>0a>0, then the value of aaa is: