The area bounded by \( 2x+y=8,\ y=a \) and \( y=a+2 \) with y-axis is 5 sq.units where \( a\gt0 \), then the value of \( a \) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
The region lies between the horizontal lines \( y=a \) and \( y=a+2 \), to the right of the y-axis and to the left of the line. From \( 2x+y=8 \), \( x=4-\frac{y}{2} \).
\( \displaystyle A=\int_{a}^{a+2}\left(4-\frac{y}{2}\right)dy \)
\( \displaystyle =\left[4y-\frac{y^{2}}{4}\right]_{a}^{a+2} \)
\( \displaystyle =8-\frac{(a+2)^{2}-a^{2}}{4} \)
\( \displaystyle =8-(a+1) \)
\( \displaystyle =7-a \)
The area is 5, so
\( \displaystyle 7-a=5 \)
\( \displaystyle \Rightarrow\ a=2 \)
Shortcut: The region is a trapezium of height 2 whose parallel sides are the x-values at \( y=a \) and \( y=a+2 \), namely \( 4-\frac{a}{2} \) and \( 3-\frac{a}{2} \). Its area is
\( \displaystyle \frac{1}{2}(7-a)\times2=7-a \).