# Calculus

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1) Find the derivative with respect to x of

Integral from 1 to X^3 of sqrt( 1 + t²) dt

I got sqrt( 1 + X^6) * 3x². Is that right?

2) Integral from 0 to pi/2 of

Xcos(X²)dx

• Calculus -

For these kind of integrals you either have to go to tables of integrals or find a suitable computer program

I found the integral of √(1+t^2) to be
[ln(√(1+t^2) + t)]/2 + [t√(1+t^2)]/2

after evaluating this from 1 to x^3 I got

[ln(√(1+x^6) + x^3)]/2 + [x^3√(1+x^6)]/2 - [ln(√2 + 1) + √2]/2

now you have to differentiate, ughhh!
Somehow I don't think this would simplify down to √( 1 + x^6) * 3x²

integral of x(cos(x^2))dx from 0 to pi/2

= (1/2)sin x^2 │ from 0 to pi/2
= 1/2 sin [(pi/2)^2] - 1/2 sin 0
= .312133

• Calculus -

The answer to question 1) is indeed
sqrt( 1 + X^6) * 3x².

You can put x^3 = y and differentiate w.r.t. y and then multiply by the derivative of y w.r.t. x (chain rule). The derivative w.r.t. y is sqrt(1+y^2) =sqrt(1+x^6) and the derivative of y w.r.t. x is 3 x^2.

• Calculus -

Let's think for a minute about what this first question means.
We have some function y(t) which we integrate from t = 1 to t = x^3.
Then we want to find out how much the area under the function changes for a smal change in x.
Well the change of the area for a change dx in x is in fact the value of the function at t=x^3
Graph the function y = f(t) (any old function, straight line will do.
Look at the area under the function from t = 1 to t = x^3
look at how that area changes for a small change in t

f(t)dt when t = x^3 is:

sqrt(1+x^6) (3 x^2)

because
f(t) = sqrt(1+t^2)
and
dt = 3 x^2 dx

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