Can someone please explain the steps on how to find the area under the curve from x=0 to x=2. When the function y= x/x^2+1.
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$\displaystyle \frac{x}{x^2+1}$ or $\displaystyle \frac{x}{x^2}+1$?
(this one)
Let $\displaystyle u=x^2+1$ $\displaystyle du=2xdx$ $\displaystyle \frac {1}{2}du=xdx$ $\displaystyle \frac {1}{2} \int_0^2 \frac {du} {u}$ $\displaystyle \frac {1}{2} \bigg [\ln {x^2+1} \bigg ]_0^2$
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