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Calculus/Integration

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Expert: - 4/2/2009


Question
{ <- integral sign

d/dx {sin(t^2) dt
upper limit -> x^3
lower limit -> x

a. sin(x^6) - sin(x^2)
b. 6x^2sin(x^3) - 2sinx
c. 3x^2(sin(x^6)) - sin(x^2)
d. 6x^5(sin (x^6)) - 2sin(x^2)
e. 2x^3(cos(x^6)) - 2x(cos(x^2))

For this problem i don't get why there's a d/dx out in the front, and how to take the integral of (sin t^2).  If you could show some of the steps to solving this, it would be very much appreciated

Answer
The actual integral sign is
⌠, alt-244, and
⌡, alt-245.

The derivative is the opposite of the integral.

All they do is cancel out.

The answer is the value of the function at the upper limit -
the value of the function at the lower limit.

That value at the upper limit is sin((x^3)²), which sin(x^6).
The value at the lower limit is sin(x²).

The answer is the one which has the difference between those two values.

Note that there are no factors out front on either answer, so it's not b, c, d or e.

Calculus

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