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Example
Created on 2024-06-20Asked by Aiden Johnson (Solvelet student)
Evaluate the integral (3x2+2x+1)dx \int (3x^2 + 2x + 1) \, dx .

Solution

To evaluate the integral (3x2+2x+1)dx \int (3x^2 + 2x + 1) \, dx : Integrate each term separately: 3x2dx+2xdx+1dx \int 3x^2 \, dx + \int 2x \, dx + \int 1 \, dx Use the power rule for integration: xndx=xn+1n+1+C \int x^n \, dx = \frac{x^{n+1}}{n+1} + C Apply the power rule to each term: 3x2dx=3x33=x3 \int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3 2xdx=2x22=x2 \int 2x \, dx = 2 \cdot \frac{x^2}{2} = x^2 1dx=x \int 1 \, dx = x Combine the results: (3x2+2x+1)dx=x3+x2+x+C \int (3x^2 + 2x + 1) \, dx = x^3 + x^2 + x + C Therefore, the integral (3x2+2x+1)dx \int (3x^2 + 2x + 1) \, dx is: (3x2+2x+1)dx=x3+x2+x+C \int (3x^2 + 2x + 1) \, dx = x^3 + x^2 + x + C Where C C is the constant of integration. Solved on Solvelet with Basic AI Model
Some of the related questions asked by Olivia Mitchell on Solvelet
1. Compute the indefinite integral (3x2+2x1)dx\int (3x^2 + 2x - 1) \, dx,2. Compute the integral x2exdx\int x^2 e^x \, dx using integration by parts.,
DefinitionWe know that Integration to give us anti-derivative of some functions, it's a mathematical prcocess which relates to Small thing concept,we can say the sum of example Area under curve for a closed loop of function, Length of curve from point A to point B, volume of revolution of a curve over certain domain represented as Integration over the varied interval. You must also be familiar with the two types of integration which are; the Indefinite and Definite integral. A definite integral represents a value over an interval, an indefinite integral represents a whole family of antiderivatives. Calculus is the study of change, and integration is the flip side of differentiation (long division) used in calculus to take an integral. Example:∫x2dx=3x3​+Cwhere C=cintegration constant.
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