Question

Use the Theorem of Calculus to calculate

int0 1 [x.sin(x2)].dx

Answer 1

First, we need to find the anti-derivative of the function f(x) = x.sin(x2), ***ie***, calculate the integral:

int  [x.sin(x2)].dx

Let's change the variables:

u = x2 => du/dx = 2.x

Which implies du/2 = x.dx

Substituting in the integral, we get:

int [x.sin(x2)].dx = int [sin(x2).x].dx =

= int [sin(u).du/2]

Using the property:

int [k.g(x)]dx = k. int [g(x)]

we have:


int [x.sin(x2)].dx = 1/2 × int[sin(u)].du

Using the known integral:

1/2 × [-cos(u)] + C

ie

F(x) = -cos(u)/2 + C

Then, F is Such that F'(x) = f(x)

Now, we just calculate in interval [0,1], ie

[F(x)]0 1 = F(1) - F(0)

= -cos(1)/2  + C - [-cos(0)/2 + C]

Simplifying:


int 0 1 [x.sin(x^2)].dx ==  [-cos(1) - 1]/2 

=  [-cos(1) - 1]/2