The value of the integral \(\mathop \smallint \limits_{ - \infty }^\infty \frac{{dx}}{{1 + {x^2}}}\) is
π
The question asks for the value of the definite integral of the function \(\frac{1}{{1 + {x^2}}}\) over the entire real line, from negative infinity to positive infinity. This is a common integral in calculus.
The integral we need to evaluate is given by: \[\mathop \smallint \limits_{ - \infty }^\infty \frac{{dx}}{{1 + {x^2}}}\]
To solve this definite integral, we first need to find the antiderivative of the integrand \(\frac{1}{{1 + {x^2}}}\).
Now, we evaluate the definite integral using the fundamental theorem of calculus for improper integrals. This involves taking limits as \(x\) approaches infinity and negative infinity.
The evaluation proceeds as follows: \[\mathop \smallint \limits_{ - \infty }^\infty \frac{{dx}}{{1 + {x^2}}} = \left[ {\arctan(x)} \right]_{ - \infty }^\infty \] This can be written as: \[\lim_{b \to \infty} \arctan(b) - \lim_{a \to -\infty} \arctan(a)\]
Substitute these limit values back into the expression for the definite integral: \[\frac{\pi}{2} - \left( -\frac{\pi}{2} \right)\] \[= \frac{\pi}{2} + \frac{\pi}{2}\] \[= \pi\]
Thus, the value of the integral \(\mathop \smallint \limits_{ - \infty }^\infty \frac{{dx}}{{1 + {x^2}}}\) is \(\pi\). This integral is a classic result in calculus, often associated with the calculation of areas related to the Cauchy distribution or probability.
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