Solve for the definite integral. [tex]$\int_1^0 \frac{1}{2 x+1} d x$[/tex]
Answer (2)
The definite integral ∫ 1 0 2 x + 1 1 d x evaluates to − 2 1 ln ( 3 ) . This result is derived from finding the antiderivative of the integrand and applying the limits of integration. The numerical approximation of this value is approximately -0.5493.
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Find the antiderivative of 2 x + 1 1 , which is 2 1 ln ∣2 x + 1∣ .
Evaluate the antiderivative at the limits of integration: 2 1 ln ∣2 ( 0 ) + 1∣ − 2 1 ln ∣2 ( 1 ) + 1∣ .
Simplify the expression: 2 1 ln ( 1 ) − 2 1 ln ( 3 ) = 0 − 2 1 ln ( 3 ) .
The final answer is − 2 1 ln ( 3 ) .
Explanation
Problem Setup We are asked to evaluate the definite integral ∫ 1 0 2 x + 1 1 d x .
Finding the Antiderivative First, we find the antiderivative of the integrand 2 x + 1 1 . Using the substitution method, let u = 2 x + 1 , so d x d u = 2 , and d x = 2 1 d u . Thus, the integral becomes ∫ 2 x + 1 1 d x = ∫ u 1 ⋅ 2 1 d u = 2 1 ∫ u 1 d u = 2 1 ln ∣ u ∣ + C = 2 1 ln ∣2 x + 1∣ + C .
Evaluating the Definite Integral Now, we evaluate the definite integral using the antiderivative. We have ∫ 1 0 2 x + 1 1 d x = [ 2 1 ln ∣2 x + 1∣ ] 1 0 = 2 1 ln ∣2 ( 0 ) + 1∣ − 2 1 ln ∣2 ( 1 ) + 1∣ = 2 1 ln ( 1 ) − 2 1 ln ( 3 ) = 0 − 2 1 ln ( 3 ) = − 2 1 ln ( 3 ) .
Final Calculation The value of the definite integral is − 2 1 ln ( 3 ) . We can approximate this value as − 2 1 ln ( 3 ) ≈ − 0.5493.
Examples
Definite integrals are used to calculate areas under curves, which has applications in physics (calculating displacement from velocity), engineering (calculating work done by a force), and economics (calculating consumer surplus). For example, if you want to find the total revenue generated by a product over a certain period, you can integrate the revenue function over that period.
