√³ — ( y = x²) e³ ³ d x = У 8 二記 e -Y do y²=x² У = ge²* ((y² - ≤3³y) (y² - 3y²-y)) =&e³·(g³- £3³ +3² + ≤3³) -y = √18 e²³. (zy³) ニキビ 3 iy³

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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The answer is 1/6 * e^-y * y^3.  Please tell me which steps are wrong in my work, thank you!

√³ — ( y = x²) e³ ³ d x
=
У
8
二記
e
-Y
do
y²=x²
У
= ge²* ((y² - ≤3³y) (y² - 3y²-y))
=&e³·(g³- £3³ +3² + ≤3³)
-y
= √18 e²³. (zy³)
ニキビ
3
iy³
Transcribed Image Text:√³ — ( y = x²) e³ ³ d x = У 8 二記 e -Y do y²=x² У = ge²* ((y² - ≤3³y) (y² - 3y²-y)) =&e³·(g³- £3³ +3² + ≤3³) -y = √18 e²³. (zy³) ニキビ 3 iy³
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