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Check out a sample textbook solutionChapter 7 Solutions
Calculus: Early Transcendentals (2nd Edition)
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- a b and the other image please is the same thank you!arrow_forwardGaussian functions G.1 G.2 G.3 G.4 G.5 G.6 G.7 G.8 8 S 8 0 e-ax² 8 1 [["new" daw ! dx= 0 2a dx= 8 [x²e-dx-1(5) x²e-ax² dx = 4 1 T erfz= 8 S™ 0 11 S x³e-ax² dx= 0 0 2 a 2 3 [x²e-max->()" x¹e-ax² dx= 1 2q² So 1/2 1/2 e-x² dx 1/2 TC¹ m! x² ₂2m+1e¬ax² dx = 2qm+1 x²me-ax² dx = 1/2 erfc z=1-erf z (2m-1)!! T 2m+1 am (2m-1)!!=1×3×5...×(2m-1) (1) a 1/2arrow_forwardPls helparrow_forward
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