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$$\int_0^1\int_x^{\sqrt{x}}\frac{e^y}{y}\,dy\,dx$$

The main problem for this is I don't know how to integrate $\dfrac{e^y}{y}$, can anyone help me out?

  • 4
    Do u know how to change order of integration? – 2017-02-04
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    The integral really simplifies. This trick it's new for me. Nice. – 2017-02-04

1 Answers 1

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This is the region over wich the integration runs: the points between the straight line $y=x$ and the arc of the parabola $y=\sqrt x$. Putting $x$ as function of $y$, the limits are $x=y^2$ and $x=y$. The intersections are at $(0,0)$ and $(1,1)$

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$$\int_0^1\int_x^{\sqrt{x}}\frac{e^y}{y}\,dy\,dx=\int_0^1\int_{y^2}^y\frac{e^y}{y}\,dx\,dy=$$

$$\int_0^1\left[x\frac{e^y}{y}\right]_{y^2}^y\,dy=\int_0^1\left(y\frac{e^y}{y}-y^2\frac{e^y}{y}\right)\,dy=$$

$$=\int_0^1\left(e^y-ye^y\right)\,dy=\left[e^y-(y-1)e^y\right]_0^1=e-2$$