Let f(x) = ∫uxvxgtdt. Then, f'(x) is equal to from Math

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 Multiple Choice QuestionsMultiple Choice Questions

571.

0π2xsinx . cosxcos4x + sin4xdx is equal to

  • π28

  • π216

  • 1

  • 0


572.

Use Simpson's 13 rule to find the value of 15fxdx given,
x 1 2 3 4 5
y 10 50 70 80 100

  • 140.88

  • 256.66

  • 160.26

  • None of these


573.

x + sinx1 + cosxdx is equal to

  • xlog1 + cosx + c

  • 1xlog1 + cosx + c

  • xtanx2 + c

  • x2tan-1x2 + c


574.

The value of cosxxdx will be

  • 2sinx + c

  • 2cosx + c

  • 2sinx + c

  • 2sinx + c


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575.

The value of 23x5 - x + xdx will be

  • 32

  • 12

  • 12

  • 13


576.

The integral 110x3dx is approximately evaluated by Trapezoidal rule 110x3dx = 31 + 1032 + α + 73 for  n = 3, then the value of α is

  • 43

  • 42

  • 53

  • None of these


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577.

Let f(x) = uxvxgtdt. Then, f'(x) is equal to

  • g[v(x)] - g[u(x)]

  • g'[v(x)] - g'[u(x)]

  • g[v(x)] v'(x) - g[u(x)] u'(x)

  • None of the above


C.

g[v(x)] v'(x) - g[u(x)] u'(x)

fx = uxvxgtdtDifferentiating with respect to x on both sides and apply Leibnitz-Rule,ddxfx = ddxuxvxgtdt    f'x = gvxddxvx - guxddxux    f'x = gvx . v'x - gux . u'x


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578.

The value of 0πcosxdx is

  • 1

  • 2

  • 0

  • 3


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579.

sinx + cosx1 + sin2xdx equals

  • logsinx + cosx

  • x

  • logsincosx

  • logx


580.

Using trapezoidal rule and taking n = 4, the value of 02dx1 + x will be

  • 1.1167

  • 1.1176

  • 1.118

  • None of these


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