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Exercise 5 Normal. , e-x² dx = \TT -00 Use 1? = (Se-x* dx)( S,e-y² dy) = L. Le-(x²+y²) dx dy = Sp2 e-(x²+y?) d(x,y) r d(r,0), %3D -0- using chg of variable (x, y) = (r cos 0, r sin 0) = g¬'(r,8) = with d(x, y) = det( d(r,6) = r d(r,8) S s re-r* dr de = n. -2n a(r,0)

Exercise 5 Normal. , e-x² dx = \TT -00 Use 1? = (Se-x* dx)( S,e-y² dy) = L. Le-(x²+y²) dx dy = Sp2 e-(x²+y?) d(x,y) r d(r,0), %3D -0- using chg of variable (x, y) = (r cos 0, r sin 0) = g¬'(r,8) = with d(x, y) = det( d(r,6) = r d(r,8) S s re-r* dr de = n. -2n a(r,0)

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter4: Calculating The Derivative
Section4.1: Techniques For Finding Derivatives
Problem 25E: Explain the relationship between the slope and the derivative of fx at x=a.
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Defn. f(x) = exp{-x-k)², 202 1(x € R') is a N([mean =] µ, [var =] o²) PDF. V2no? c) re) = Vĩ. Hint: Vī = Le-x* dx = 2 ° e¯x* dx. Now use chg-of-variable y = x2. = VT.
Transcribed Image Text:Defn. f(x) = exp{-x-k)², 202 1(x € R') is a N([mean =] µ, [var =] o²) PDF. V2no? c) re) = Vĩ. Hint: Vī = Le-x* dx = 2 ° e¯x* dx. Now use chg-of-variable y = x2. = VT.
Exercise 5 Normal. , e-x² dx = \TT -00 * dy) = L. Se-(x²+y?) dx dy = Sp2 e-(x²+y*) d(x,y) r d(r,0), Use 1? = (Se-x* dx)( Le-y² dy) = S Le-(x²+y?) dx dy = Sp2 e-(x²+y²) d(x,y) -0- using chg of variable (x, y) = (r cos 0, r sin 0) = g¬'(r,8) ag-1(r.0) = with d(x, y) = |det( d(r,0) = r d(r,0) S s re-r* dr do = n. 27n
Transcribed Image Text:Exercise 5 Normal. , e-x² dx = \TT -00 * dy) = L. Se-(x²+y?) dx dy = Sp2 e-(x²+y*) d(x,y) r d(r,0), Use 1? = (Se-x* dx)( Le-y² dy) = S Le-(x²+y?) dx dy = Sp2 e-(x²+y²) d(x,y) -0- using chg of variable (x, y) = (r cos 0, r sin 0) = g¬'(r,8) ag-1(r.0) = with d(x, y) = |det( d(r,0) = r d(r,0) S s re-r* dr do = n. 27n
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