For A>0, compute I(A)=∫-A^A (d t)/(1+t^2) and evaluate lima →∞ I(A), the area under the

step 1 So we notice that the function inside here is the even function. So here we can write the integr

For A>0, compute I(A)=∫-A^A (d t)/(1+t^2) and...

For $\quad A>0, \quad$ compute $\quad I(A)=\int_{-A}^{A} \frac{d t}{1+t^{2}} \quad$ and evaluate $\lim _{a \rightarrow \infty} I(A),$ the area under the graph of $\frac{1}{1+t^{2}}$ on $[-\infty, \infty]$.

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Key Concepts

Antiderivative of 1/(1+t^2)

The function 1/(1+t^2) has an elementary antiderivative given by the arctangent function, arctan(t). This integral result is a standard example in calculus, and knowing this antiderivative is crucial to evaluating the definite integral over any interval.

Symmetry in Integration

Due to the even symmetry of the function 1/(1+t^2), integration over symmetric limits (from -A to A) can be simplified. Recognizing symmetry helps reduce computational complexity and provides insight into how the antiderivative behaves at symmetric endpoints, where arctan(-A) is the negative of arctan(A).

Improper Integrals and Convergence

Improper integrals extend the idea of integration to intervals that are unbounded, or to functions with discontinuities. Evaluating the limit as A approaches infinity in the definite integral tests whether the total area under the curve converges to a finite number. This concept is essential in determining whether an infinite interval yields a finite area.

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