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The area (in sq.units) bounded by the lines y=2x+3y=2x+3, y=0y=0 between x=−2x=-2 and x=1x=1 is

Solution

✅ Correct Option: 4

The line meets y=0y=0 at x=−32x = -\dfrac{3}{2}, so it lies below the axis on [−2,−3/2][-2,-3/2] and above on [−3/2,1][-3/2,1].

Area =∣∫−2−3/2(2x+3)dx∣+∫−3/21(2x+3)dx=14+254=264=132= \left|\int_{-2}^{-3/2}(2x+3)dx\right| + \int_{-3/2}^{1}(2x+3)dx = \dfrac{1}{4} + \dfrac{25}{4} = \dfrac{26}{4} = \dfrac{13}{2}.

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