Question

23. \int_{1}^{e} \int_{1}^{e} \frac{\ln x}{xy} dy dx \newline (See Problem 13.) \newline 24. \int_{1}^{e^{2}} \int_{1}^{e} \frac{\ln x}{xy} dx dy \newline (See Problem 14.) \newline 25. \int_{0}^{2} \int_{0}^{1} 3y^{2}e^{x+y^{3}} dx dy \newline (See Problem 15.)

Name: 23. ∫1^e ∫1^e (lnx)/(xy) dy dx (See Problem 13.) 24. ∫1^e^2 ∫1^e (lnx)/(xy) dx dy (See Problem 14.) 25. ∫0^2 ∫0^1 3y^2e^x+y^3 dx dy (See Problem 15.)
Uploaded: 2022-06-24T12:27:24-08:00
Duration: 5 min 22 s
Channel: Madhur L
Description: 23. ∫1^e ∫1^e (lnx)/(xy) dy dx (See Problem 13.) 24. ∫1^e^2 ∫1^e (lnx)/(xy) dx dy (See Problem 14.) 25. ∫0^2 ∫0^1 3y^2e^x+y^3 dx dy (See Problem 15.)

          23. \int_{1}^{e} \int_{1}^{e} \frac{\ln x}{xy} dy dx \newline (See Problem 13.) \newline 24. \int_{1}^{e^{2}} \int_{1}^{e} \frac{\ln x}{xy} dx dy \newline (See Problem 14.) \newline 25. \int_{0}^{2} \int_{0}^{1} 3y^{2}e^{x+y^{3}} dx dy \newline (See Problem 15.)

23. ∫1^e ∫1^e (lnx)/(xy) dy dx (See Problem 13.) 24. ∫1^e^2 ∫1^e (lnx)/(xy) dx dy (See Problem 14.) 25. ∫0^2 ∫0^1 3y^2e^x+y^3 dx dy (See Problem 15.)

Added by Kathy T.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

06/24/2022

Step 1

We are given the integral: (A) \int 3y^2 e^(x+y^3) dx To find the antiderivative, we can use the substitution method. Let u = x + y^3. Then, du/dx = 1, and we can write dx = du. Now, substitute u into the integral: \int 3y^2 e^u du Since the integrand does not Show more…

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In Problems 7-16, find each antiderivative. Then use the antiderivative to evaluate the definite integral. 15. (A) int 3y^2 e^(x+y^3) dx (B) int_0^1 3y^2 e^(x+y^3) dx In Problems 17-26, evaluate each iterated integral. (See the indicated problem for the evaluation of the inner integral.) 23. int_1^2 int_1^2 (lnx)/(xy) dydx (See Problem 13.) 25. int_0^2 int_0^1 3y^2 e^(x+y^3) dxdy (See Problem 15.)