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D e a r S t u d e n t, \int_{1}^{4} \sqrt{x} d x \Rightarrow \int_{1}^{4} x^{\frac{1}{2}} d x = {[\frac{x^{\frac{1}{2} + 1}}{^{\frac{1}{2} + 1}}]}^{4}_{1} = \frac{2}{3} \times {[x^{\frac{3}{2}}]}^{4}_{1} = \frac{2}{3} \times {[{(\sqrt{x})}^{3}]}^{4}_{1} \frac{2}{3} \times [2^{3} - 1] = \frac{2}{3} \times [7] = \frac{14}{3}, O p t i o n 1 R e g a r d s

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