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
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
for your second question
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
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.
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
Now for your case
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
To find the derivative of the integral in the first question, you'll need to use the Fundamental Theorem of Calculus and apply the chain rule. Here's the step-by-step process:
Step 1: Rewrite the integral as a function of x, using the variable of integration t:
∫[1 to x^3] √(1 + t²) dt
Step 2: Let F(x) be the antiderivative of f(t) = √(1 + t²). We know that:
d/dx ∫[a to g(x)] f(t) dt = g'(x) * f(g(x))
Step 3: Apply the Fundamental Theorem of Calculus to differentiate the integral with respect to x:
d/dx ∫[1 to x^3] √(1 + t²) dt = d/dx [ F(x^3) ]
Step 4: Apply the chain rule to the right-hand side:
d/dx [ F(x^3) ] = F'(x^3) * (d/dx [x^3])
Step 5: Differentiate the function F(x) using the chain rule:
F'(x) = √(1 + x²)
Step 6: Differentiate the term (x^3) with respect to x:
(d/dx [x^3]) = 3x^2
Putting everything together, we have:
d/dx [ F(x^3) ] = F'(x^3) * (d/dx [x^3]) = √(1 + x^6) * 3x^2
So your previous answer of sqrt( 1 + X^6) * 3x^2 is correct!
Moving on to the second question:
To evaluate the given integral, you need to use integration techniques. Here's the step-by-step process:
Step 1: Rewrite the integral:
∫[0 to π/2] xcos(x²) dx
Step 2: Let u = x²:
du/dx = 2x => dx = du / (2x)
Step 3: Substitute the new variable and differential into the integral:
∫[0 to π/2] (1/2)(u^(1/2) * cos(u)) du
Step 4: Simplify the integral:
(1/2) ∫[0 to π/2] u^(1/2) * cos(u) du
Step 5: Integrate using integration techniques (such as integration by parts or substitution). Unfortunately, there is no simple closed-form solution for this type of integral.
Therefore, the integral ∫[0 to π/2] xcos(x²) dx does not have an elementary solution and cannot be evaluated precisely in terms of elementary functions. However, you can approximate the value using numerical methods or software.