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The area of the region bounded by the lines x=2y+3,x=0,y=1x = 2y + 3, x = 0, y = 1 and y=−1y = -1 is:

Solution

✅ Correct Option: 2

Area =∫−11(2y+3) dy=[y2+3y]−11=(1+3)−(1−3)=4−(−2)=6= \int_{-1}^{1} (2y+3)\,dy = [y^2 + 3y]_{-1}^{1} = (1+3) - (1-3) = 4 - (-2) = 6 sq. units.

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