Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
Step-by-step explanation:
Given: [tex]f'(x) = x^2e^{2x^3}[/tex] and [tex]f(0) = 0[/tex]
We can solve for f(x) by writing
[tex]\displaystyle f(x) = \int f'(x)dx=\int x^2e^{2x^3}dx[/tex]
Let [tex]u = 2x^3[/tex]
[tex]\:\:\:\:du=6x^2dx[/tex]
Then
[tex]\displaystyle f(x) = \int x^2e^{2x^3}dx = \dfrac{1}{6}\int e^u du[/tex]
[tex]\displaystyle \:\:\:\:\:\:\:=\frac{1}{6}e^{2x^3} + k[/tex]
We know that f(0) = 0 so we can find the value for k:
[tex]f(0) = \frac{1}{6}(1) + k \Rightarrow k = -\frac{1}{6}[/tex]
Therefore,
[tex]\displaystyle f(x) = \frac{1}{6} \left(e^{2x^3} - 1 \right)[/tex]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.