Calculus help?

1 ⤊

Find the maximum value of d/dx[integralcosxdx]

Thanks

2006-12-08 04:08:31 · 4 answers · asked by magt_g 1 in Science & Mathematics ➔ Mathematics

4 answers

the max value is = 1

notice that d/dx[integralcosxdx] = cosx .

2006-12-08 04:21:04 · answer #1 · answered by Anonymous · 1⤊ 0⤋

Find the maximum value of d/dx[integralcosxdx]
[integralcosxdx] = sinx +C
Thus d/dx[integralcosxdx] d/dx (sinx +C) = cos x
Max value occurs x = 0, 2pi, ...... 2npi and is eqal to 1

2006-12-08 12:21:46 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋

d/dx[â«cos x dx] = cos x

The maximum value of cos x is 1.

₢

2006-12-08 12:37:13 · answer #3 · answered by Luiz S 7 · 0⤊ 0⤋

integral=cosxdx
on integrating this is sinx
maximum value is 1

2006-12-08 12:10:26 · answer #4 · answered by raj 7 · 0⤊ 0⤋