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- Thread starter MarkFL
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- Feb 7, 2012

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Spot on as always.

I used essentially the same method:

He had already determined the equation of the tangent line to be:

\(\displaystyle y=5x+2\)

To see that this is true, we find by differentiating $f$ with respect to $x$:

\(\displaystyle f'(x)=\left(e^x+x \right)e^x+\left(e^x+1 \right)^2\)

Hence, we find:

\(\displaystyle f'(0)=5\)

and then applying the point-slope formula, we have:

\(\displaystyle y-2=5(x-0)\)

\(\displaystyle y=5x+2\)

And so the area under the tangent line and bounded by the axes is a right triangle whose base is the magnitude of the $x$-intercept, or $\dfrac{2}{5}$, and whose altitude is $2$, which means the area under the line is:

\(\displaystyle A_T=\frac{1}{2}\cdot\frac{2}{5}\cdot2=\frac{2}{5}\)

To find the area under the curve, I instructed him to let $x_0$ represent this root, and so we may write:

\(\displaystyle A_C=\int_{x_0}^0 \left(e^x+x \right)\left(e^x+1 \right)\,dx\)

Next, use the substitution:

\(\displaystyle u=e^x+x\,\therefore\,du=\left(e^x+1 \right)\,dx\)

Since $f\left(x_0 \right)=0$, then we must have $e^{x_0}+x_0=0$ as well, because the other factor $e^x+1$ has no real roots. And so our definite integral becomes:

\(\displaystyle A_C=\int_0^1 u\,du=\frac{1}{2}\left[u^2 \right]_0^1=\frac{1}{2}\)

Finally, since the shaded area $A$ is the area under the curve less the area under the tangent line, we may write:

\(\displaystyle A=A_C-A_T=\frac{1}{2}-\frac{2}{5}=\frac{1}{10}\)