Why do equal triangular areas above and below the axis result in a zero definite integral over that interval?

The ordinary integral $\int_{-\infty}^{\infty} g(t)e^{-2\pi i ft} dt$ is well-defined if $g(t)$ is absolutely integrable (i.e., $\int |g(t)| dt < \infty$). However, ideal sinusoids sustained indefinitely are not absolutely integrable because their energy spreads over infinite time.

Conditions: Considering the limit as the time window $T \to \infty$.; Analyzing either decaying transient signals or sustained periodic signals.

The new integration bounds are $x=2$ and $x=3$. When $u=0$, $x = 1 + e^0 = 1 + 1 = 2$.

Conditions: The substitution is $x = 1 + e^u$.; The original bounds are $u=0$ and $u=\ln 2$.; The exponential function $e^u$ is evaluated at these points.

The definite integral can be evaluated by recognizing that the integrand $y = \sqrt{9-x^2}$ graphs as an upper semicircle of radius 3. Because the function is continuous and nonnegative on $[-3, 3]$, the integral equals the ordinary geometric area of this region.

Conditions: The integrand is recognized as the upper semicircle of $x^2+y^2=9$.; The function is continuous and nonnegative on the interval $[-3, 3]$.; Use the real geometric area formula for a circle.

The equation $x^2 + y^2 = 9$ describes a full circle centered at the origin with radius 3, which includes both upper and lower branches. However, the original function is defined using the principal square root, $y = \sqrt{9-x^2}$.

Conditions: Working over the real numbers.; Using the principal (nonnegative) square root convention.; The underlying relation is the circle $x^2 + y^2 = 9$.

To evaluate the definite integral, substitute $x = 1 + e^u$, which gives $dx = e^u du$. Convert the bounds from $u=0$ and $u=\ln 2$ to $x=2$ and $x=3$.

Conditions: The integral is definite with bounds $u=0$ to $u=\ln 2$.; The integrand is $e^u \sqrt{1+e^u}$.; Substitution $x = 1 + e^u$ is used.