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

Identify Integral: The integral we want to solve is ∫ from 0 to 1 of 1/(1+x^2) dx. The antiderivative of 1/(1+x^2) is known to be arctan(x), because the derivative of arctan(x) is 1/(1+x^2).Apply Fundamental Theorem: We will apply the Fundamental Theorem of Calculus, which states that if F is an antiderivative of f on an interval [a, b], then ∫ from a to b of f(x) dx = F(b) - F(a).Substitute Limits: We substitute x with the upper limit of the integral and then with the lower limit of the integral and find the difference. So we calculate arctan(1) - arctan(0).Calculate Values: arctan(1) is π/4 because tan(π/4) = 1. And arctan(0) is 0 because tan(0) = 0.Final Result: Subtracting these two values gives us π/4 - 0, which simplifies to π/4.

Home

Math Problems

Calculus

Find derivatives of logarithmic functions

Full solution

Q. Integrate

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

for a limit

[0,1]

Identify Integral: The integral we want to solve is $\int_{0}^{1} \frac{1}{1+x^2} \, dx$ . The antiderivative of $\frac{1}{1+x^2}$ is known to be $\arctan(x)$ , because the derivative of $\arctan(x)$ is $\frac{1}{1+x^2}$ .
Apply Fundamental Theorem: We will apply the Fundamental Theorem of Calculus, which states that if $F$ is an antiderivative of $f$ on an interval $[a, b]$ , then $\int_{a}^{b} f(x) \, dx = F(b) - F(a)$ .
Substitute Limits: We substitute $x$ with the upper limit of the integral and then with the lower limit of the integral and find the difference. So we calculate $\arctan(1) - \arctan(0)$ .
Calculate Values: $\text{arctan}(1)$ is $\frac{\pi}{4}$ because $\tan\left(\frac{\pi}{4}\right) = 1$ . And $\text{arctan}(0)$ is $0$ because $\tan(0) = 0$ .
Final Result: Subtracting these two values gives us $\pi/4 - 0$ , which simplifies to $\pi/4$ .