Find the area bounded by $y = x^{2} + 2$ above x-axis between $x = 2$ & $x = 3$

\frac{25}{6} s q . u n i t s

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\frac{25}{3} s q . u n i t s

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\frac{25}{2} s q . u n i t s

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\frac{25}{12} s q . u n i t s

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Solution

The correct option is B $\frac{25}{3} s q . u n i t s$
Required are $= \int_{2}^{3} (x^{2} + 2) d x = {[\frac{x^{3}}{3} + 2 x]}_{2}^{3} = 15 - \frac{20}{3} = \frac{25}{3}$ sq. units

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Find the area bounded by y=x2+2 above x-axis between x=2 & x=3

The correct option is B 253sq.unitsRequired are =∫32(x2+2)dx=[x33+2x]32=15−203=253 sq. units

Find the area bounded by $y = x^{2} + 2$ above x-axis between $x = 2$ & $x = 3$

The correct option is B $\frac{25}{3} s q . u n i t s$
Required are $= \int_{2}^{3} (x^{2} + 2) d x = {[\frac{x^{3}}{3} + 2 x]}_{2}^{3} = 15 - \frac{20}{3} = \frac{25}{3}$ sq. units