Asked by Vanessa
Evaluate the definite integral.
x sqrt(13 x^2 + 36)dx between (0,1)
x sqrt(13 x^2 + 36)dx between (0,1)
Answers
Answered by
Kuai
x √(13x^2 +36)dx
u = 13x^2 + 36
between (36, 49)
du= 26x
du/26 = x
1/26 (u)^1/2 du
1/26 *2/3 (u)^3/2 to 36 to 49
1/39[ 49^3/2 -36^3/2]
1/39( 343-216) = 127/39
u = 13x^2 + 36
between (36, 49)
du= 26x
du/26 = x
1/26 (u)^1/2 du
1/26 *2/3 (u)^3/2 to 36 to 49
1/39[ 49^3/2 -36^3/2]
1/39( 343-216) = 127/39
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