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Improper Integrals Calculator

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Example
Created on 2024-06-20Asked by Avery Moore (Solvelet student)
Evaluate the improper integral ∫1∞1x2 dx \int_{1}^{\infty} \frac{1}{x^2} \, dx .

Solution

To evaluate the improper integral ∫1∞1x2 dx \int_{1}^{\infty} \frac{1}{x^2} \, dx : First, express the improper integral as a limit: ∫1∞1x2 dx=lim⁡b→∞∫1b1x2 dx \int_{1}^{\infty} \frac{1}{x^2} \, dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^2} \, dx Now, compute the definite integral: ∫1b1x2 dx \int_{1}^{b} \frac{1}{x^2} \, dx Recall that: ∫1x2 dx=−1x \int \frac{1}{x^2} \, dx = -\frac{1}{x} Evaluate the definite integral: [−1x]1b=−1b−(−11)=−1b+1 \left[ -\frac{1}{x} \right]_{1}^{b} = -\frac{1}{b} - \left( -\frac{1}{1} \right) = -\frac{1}{b} + 1 Take the limit as b b approaches infinity: lim⁡b→∞(−1b+1)=0+1=1 \lim_{b \to \infty} \left( -\frac{1}{b} + 1 \right) = 0 + 1 = 1 Therefore, the value of the improper integral ∫1∞1x2 dx \int_{1}^{\infty} \frac{1}{x^2} \, dx is: ∫1∞1x2 dx=1 \int_{1}^{\infty} \frac{1}{x^2} \, dx = 1 Solved on Solvelet with Basic AI Model
Some of the related questions asked by Abigail Roberts on Solvelet
1. Evaluate the improper integral ∫1∞1x dx\int_1^\infty \frac{1}{x} \, dx,2. Compute the antiderivative of the function f(x)=2sin⁡(x)+3xf(x) = 2\sin(x) + \frac{3}{x}.,
DefinitionAn improper integral is an integral for which either or both of the limits of integration are infinite and/or the integrand is unbounded at the limits of integration. They compute to limits, where the limit process is what smooths out the behavior as well as the interval being infinite. Improper integrals are a vital extension of calculus for unbounded functions or domains and occur in probability, physics, and engineering to model distributions, forces, and other continuously varying phenomena. For example, the improper integral ∫1∞​x21​dx is evaluated as limb→∞​∫1b​x21​dx=limb→∞​[−x1​]1b​=1.
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