Evaluate the iterated integral: the integral from x = 0 to 1 of x-differential, with the inner integral of y dy taken from y = x up to y = √(1 − x²) + 1.

Step-by-step solution

1. 1

   Do the inner integral, where $x$ is a constant. The integrand is just $y$, so

   $\int_{x}^{\sqrt{1-x^{2}}+1}y\,dy=\left[\frac{y^{2}}{2}\right]_{x}^{\sqrt{1-x^{2}}+1}=\frac12\Bigl(\bigl(\sqrt{1-x^{2}}+1\bigr)^{2}-x^{2}\Bigr).$

   Before expanding, note the limits are consistent on $[0,1]$: the upper limit $\sqrt{1-x^{2}}+1$ is at least $1$ while the lower limit $x$ is at most $1$, so the region is non-degenerate.

2. 2

   Expand the square — this is where the $\sqrt{\;}$ survives.

   $\bigl(\sqrt{1-x^{2}}+1\bigr)^{2}=(1-x^{2})+2\sqrt{1-x^{2}}+1=2-x^{2}+2\sqrt{1-x^{2}}.$

   The cross term $2\sqrt{1-x^{2}}$ is the only non-polynomial piece; everything else is elementary.

3. 3

   Simplify the inner result.

   $\frac12\Bigl(2-x^{2}+2\sqrt{1-x^{2}}-x^{2}\Bigr)=\frac12\Bigl(2-2x^{2}+2\sqrt{1-x^{2}}\Bigr)=1-x^{2}+\sqrt{1-x^{2}}.$

   So the double integral has become the single integral $\displaystyle\int_{0}^{1}\bigl(1-x^{2}+\sqrt{1-x^{2}}\bigr)dx$.

4. 4

   Split and evaluate each piece. The first two are immediate:

   $\int_{0}^{1}1\,dx=1,\qquad \int_{0}^{1}x^{2}\,dx=\frac13.$

   For the third, $y=\sqrt{1-x^{2}}$ on $[0,1]$ traces the first-quadrant quarter of the unit circle, so its integral is the quarter-disc area $\pi/4$ — no trigonometric substitution needed.

5. 5

   Combine and check.

   $1-\frac13+\frac{\pi}{4}=\frac23+\frac{\pi}{4}.$

   Numerically $2/3+\pi/4=1.4520648301$, and adaptive quadrature of $1-x^{2}+\sqrt{1-x^{2}}$ over $[0,1]$ returns $1.4520648301$ — the same to ten decimals.