integral from 1 to 4 (u-2)/sqrt(u) I tried doing.. $\displaystyle (u-2)(u^{{-1/2}})$ and I ended up getting -3/4 but the answer is 2/3..
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Originally Posted by elpermic integral from 1 to 4 (u-2)/sqrt(u) I tried doing.. $\displaystyle (u-2)(u^{{-1/2}})$ and I ended up getting -3/4 but the answer is 2/3.. $\displaystyle \int_1^4\frac{u-2}{\sqrt{u}}du=\int_1^4\bigg[\sqrt{u}-\frac{2}{\sqrt{u}}\bigg]du$
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