Multivariable calculus

Double Integral of x Under y = sin x from 0 to 1

The double integral of x over the region under y = sin x with 0 <= x <= 1, in radians, equals sin 1 - cos 1, about 0.301169. The inner integral in y gives x sin x, and integration by parts turns that into sin x - x cos x, evaluated from 0 to 1.

f(x,y)=xf(x,y) = x
∫01∫0sin⁡xx dy dx=sin⁡1−cos⁡1≈0.301169\int_0^1 \int_0^{\sin x} x\,dy\,dx = \sin 1 - \cos 1 \approx 0.301169

Numerically 0.301169, confirmed by quadrature on every build.

An integrand with no y makes the inner step trivial

Because xx is constant with respect to yy, the inner integral is just the integrand times the height of the slice.

∫0sin⁡xx dy=x[y]y=0y=sin⁡x=xsin⁡x\int_0^{\sin x} x\,dy = x\left[y\right]_{y=0}^{y=\sin x} = x\sin x

On [0,1][0,1] in radians sin⁡x\sin x is positive, so the slice height is genuine and the whole region sits above the axis. What remains is a one-variable integral that needs parts, with u=xu = x and dv=sin⁡x dxdv = \sin x\,dx.

∫01xsin⁡x dx=[sin⁡x−xcos⁡x]01=sin⁡1−cos⁡1\int_0^1 x\sin x\,dx = \left[\sin x - x\cos x\right]_0^1 = \sin 1 - \cos 1

Differentiate the antiderivative to confirm it: ddx(sin⁡x−xcos⁡x)=cos⁡x−cos⁡x+xsin⁡x=xsin⁡x\frac{d}{dx}\left(\sin x - x\cos x\right) = \cos x - \cos x + x\sin x = x\sin x.

The mistake: the sign buried in integration by parts

With dv=sin⁡x dxdv = \sin x\,dx the antiderivative is v=−cos⁡xv = -\cos x, and that minus sign has to survive two more steps.

∫xsin⁡x dx=−xcos⁡x−∫(−cos⁡x) dx=−xcos⁡x+sin⁡x\int x\sin x\,dx = -x\cos x - \int \left(-\cos x\right)\,dx = -x\cos x + \sin x

Writing −xcos⁡x−sin⁡x-x\cos x - \sin x is the standard wrong answer. It evaluates to −cos⁡1−sin⁡1≈−1.382-\cos 1 - \sin 1 \approx -1.382, and a negative number cannot be right for a positive integrand over a region of positive area.

  • Radians, not degrees. In radians sin⁡1≈0.8415\sin 1 \approx 0.8415; a calculator left in degrees returns 0.017450.01745 and the answer collapses.
  • The region is not a full arch. Stopping at x=1x = 1, short of the peak at x=π2x = \frac{\pi}{2}, is deliberate.
  • Bound the answer: on this region x≤1x \le 1, so the integral cannot exceed the area, 1−cos⁡1≈0.45971 - \cos 1 \approx 0.4597.

That last bound is worth writing down on any problem of this shape. Here 0.3012<0.45970.3012 < 0.4597, so the answer clears the test.

The area and the average

The region's area is the ordinary single integral of the boundary curve.

area=∫01sin⁡x dx=1−cos⁡1≈0.459698\text{area} = \int_0^1 \sin x\,dx = 1 - \cos 1 \approx 0.459698

So the average value of xx over the region is 0.3011690.459698≈0.655\frac{0.301169}{0.459698} \approx 0.655, which is the xx-coordinate of the centroid. It sits to the right of the midpoint 0.50.5 because the region is much taller near x=1x = 1 than near x=0x = 0, so more of its area carries a large xx.

Frequently asked questions

Why is there no antiderivative work in the inner step?

Because the integrand contains no yy. Integrating a constant with respect to yy gives that constant times the length of the slice, so the entire first step reduces to xx multiplied by sin⁡x\sin x.

Is sin 1 - cos 1 the exact answer?

Yes. The decimal 0.3011690.301169 is only an approximation of it, with 11 measured in radians. There is no simpler closed form, and leaving the answer as sin⁡1−cos⁡1\sin 1 - \cos 1 is what a grader expects.