Calculate the following integral: integral from 0 to 1
Calculate the following integral: $$ \int_0^1 \int_{\sqrt{y}}^1 \frac{1}{x} e^{-y/x}\,dx\,dy. $$
Explanation of Method
Integrating $\frac1x e^{-y/x}$ directly with respect to $x$ has no elementary antiderivative, so we reverse the order first: with $y$ as the inner variable, $\frac1x e^{-y/x}$ integrates cleanly via the substitution $u=-y/x$.
- Identify the region and reverse the order.
- Integrate with respect to $y$ using $u=-y/x$.
- Integrate the resulting $1-e^{-x}$ with respect to $x$.
Step 1 · Identify the region
From $\int_0^1\int_{\sqrt y}^1 \frac1x e^{-y/x}\,dx\,dy$: $0\leqslant y\leqslant1$, $\sqrt y\leqslant x\leqslant1$.
Step 2 · Sketch the region and reverse the order
The region is bounded below by $y=0$, above by $y=1$, on the left by $x=\sqrt y$ (i.e. $y=x^2$), and on the right by $x=1$. Reversed: $0\leqslant x\leqslant1$, $0\leqslant y\leqslant x^2$.
Step 3 · Rewrite the integral
$$ I = \int_0^1\int_0^{x^2} \frac1x e^{-y/x}\,dy\,dx. $$
Step 4 · Integrate with respect to $y$
Let $u=-\frac yx$, so $du=-\frac1x\,dy$. When $y=0$, $u=0$; when $y=x^2$, $u=-x$:
$$ \int_0^{x^2} \frac1x e^{-y/x}\,dy = -\int_0^{-x} e^u\,du = -[e^u]_0^{-x} = -(e^{-x}-1) = 1-e^{-x}. $$
Step 5 · Integrate with respect to $x$
$$ I = \int_0^1 (1-e^{-x})\,dx = [x]_0^1 - [-e^{-x}]_0^1 = 1 - \left(-e^{-1}+1\right) = e^{-1} = \frac1e. $$
Additional Notes
- This is a classic case where the integral is intractable in the given order but becomes elementary once the order of integration is reversed — always worth trying when the inner antiderivative looks difficult.
Result
The solution to the integral is: