Integrate 1//(1+x^(2)) for limit [0,1].

Find Antiderivative: We need to integrate the function 1/(1+x^2) within the limits from 0 to 1. The antiderivative of 1/(1+x^2) is arctan(x), since the derivative of arctan(x) is 1/(1+x^2).Apply Definite Integral: Now we will apply the definite integral with the limits from 0 to 1 to the antiderivative we found. ∫ from 0 to 1 of 1/(1+x^2) dx = [arctan(x)] from 0 to 1Evaluate at Limits: We will evaluate the antiderivative at the upper limit and then subtract the evaluation at the lower limit. arctan(1) - arctan(0)Calculate Final Result: We know that arctan(1) is π/4 because tan(π/4) = 1, and arctan(0) is 0 because tan(0) = 0. So, arctan(1) - arctan(0) = π/4 - 0 = π/4Therefore, the integral of 1/(1+x^2) from 0 to 1 is π/4.

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Calculus

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Q. Integrate

\frac{1}{1+x^{2}}

for limit

[0,1]

Find Antiderivative: We need to integrate the function $\frac{1}{1+x^2}$ within the limits from $0$ to $1$ . The antiderivative of $\frac{1}{1+x^2}$ is $\text{arctan}(x)$ , since the derivative of $\text{arctan}(x)$ is $\frac{1}{1+x^2}$ .
Apply Definite Integral: Now we will apply the definite integral with the limits from $0$ to $1$ to the antiderivative we found. $\newline$ $\int_{0}^{1} \frac{1}{1+x^2} dx = [\arctan(x)]_{0}^{1}$
Evaluate at Limits: We will evaluate the antiderivative at the upper limit and then subtract the evaluation at the lower limit. $\newline$ $\arctan(1) - \arctan(0)$
Calculate Final Result: We know that $\text{arctan}(1)$ is $\frac{\pi}{4}$ because $\tan\left(\frac{\pi}{4}\right) = 1$ , and $\text{arctan}(0)$ is $0$ because $\tan(0) = 0$ . So, $\text{arctan}(1) - \text{arctan}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$
Calculate Final Result: We know that $\arctan(1)$ is $\frac{\pi}{4}$ because $\tan\left(\frac{\pi}{4}\right) = 1$ , and $\arctan(0)$ is $0$ because $\tan(0) = 0$ . So, $\arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$ Therefore, the integral of $\frac{1}{1+x^2}$ from $0$ to $1$ is $\frac{\pi}{4}$ .