Definite Integral – An exponential function on a finite interval – Exercise 6421 Post category:Definite Integral Post comments:0 Comments Exercise Evaluate the integral ∫0ln2e3x−1dx\int_0^{\ln 2} e^{3x-1} dx∫0ln2e3x−1dx Final Answer Show final answer ∫0ln2e3x−1dx=73e\int_0^{\ln 2} e^{3x-1} dx=\frac{7}{3e}∫0ln2e3x−1dx=3e7 Solution Coming soon… Share with Friends Read more articles Previous PostDefinite Integral – A quotient of functions on a finite interval – Exercise 6412 Next PostDefinite Integral – A rational function on a symmetric interval – Exercise 6423 You Might Also Like Definite Integral – Finding area between parabola, line and axis-x – Exercise 7024 August 21, 2019 Definite Integral – Finding area between a polynomial and a line – Exercise 7006 August 21, 2019 Definite Integral – A quotient of functions on a finite interval – Exercise 6412 July 8, 2019 Definite Integral – Finding area between two functions and an asymptote – Exercise 5492 May 25, 2019 Definite integral – area computation of a bounded domain – Exercise 6615 July 20, 2019 Definite Integral – A rational function with absolute value on symmetric interval – Exercise 6601 July 16, 2019 Leave a Reply Cancel replyCommentEnter your name or username to comment Enter your email address to comment Enter your website URL (optional) Δ
Definite Integral – A rational function with absolute value on symmetric interval – Exercise 6601 July 16, 2019
