Calculate a definite integral from another definite integral

[0kelvin]

Homework Statement

f is continuous in [-1, 1]. Calculate $\int_0^1 f(2x - 1) dx$, given that $\int_{-1}^1 f(u) du = 5$

f is continuous in [0, 4]. Calculate $\int_{-2}^2 xf(x^2) dx$

The Attempt at a Solution

I did one easier exercise where both integrals were in terms of x, a simple x - 2 = u and du = dx substitution made the trick. The other fact was that I noticed that the whole function was shifted by two units and the limits of integratition were also shifted by two units, resulting in the same value for the definite integral.

 

[Mark44]

Homework Statement

f is continuous in [-1, 1]. Calculate $\int_0^1 f(2x - 1) dx$, given that $\int_{-1}^1 f(u) du = 5$

f is continuous in [0, 4]. Calculate $\int_{-2}^2 xf(x^2) dx$

The Attempt at a Solution

I did one easier exercise where both integrals were in terms of x, a simple x - 2 = u and du = dx substitution made the trick. The other fact was that I noticed that the whole function was shifted by two units and the limits of integratition were also shifted by two units, resulting in the same value for the definite integral.

These look to me like applications of substitution. Did you try any substitutions on either integral? Also, since these are definite integrals, if you make substitutions, you'll want to change the limits of integration to values of the substituted variable.

 

[0kelvin]

$u = 2x - 1 \Rightarrow x = \frac{u+1}{2}$
$du = 2dx$

This is where I was confused:

$$\int_{u = 2.0 - 1}^{u = 2.1 - 1} f(u) \frac{du}{2}$$

$\int_{-1}^1 f(u) \frac{du}{2}$

$\frac{1}{2}\int_{-1}^1 f(u) du = \frac{5}{2}$

Had an idea about the second exercise:

$u = f(x^2)$

$du = 2xf'(x^2)dx$

Thus

$\frac{1}{2}\int_{-2}^2 u2xdx$

$\frac{1}{2}\int_{-2}^2 udu$

The rest is trivial.

 

[Mark44]

$u = 2x - 1 \Rightarrow x = \frac{u+1}{2}$
$du = 2dx$

This is where I was confused:

$$\int_{u = 2.0 - 1}^{u = 2.1 - 1} f(u) \frac{du}{2}$$

$\int_{-1}^1 f(u) \frac{du}{2}$

$\frac{1}{2}\int_{-1}^1 f(u) du = \frac{5}{2}$

Right.

Your notation needs some work, though. When you write "2.0" and "2.1" I think you intend that the "." is supposed to mean multiplication, but it looks like a decimal fraction. Use * to make this clearer.

Had an idea about the second exercise:

$u = f(x^2)$

There's a much simpler substitution.

$du = 2xf'(x^2)dx$

Thus

$\frac{1}{2}\int_{-2}^2 u2xdx$

No this won't work.

$\frac{1}{2}\int_{-2}^2 udu$

The rest is trivial.

The last integral is $\frac{1}{2}\left. \frac{f(x^2)}{2} \right|_{x = -2}^2$
But you don't have any information about that function.

 

[0kelvin]

Oh, -_-'' I just noticed that I substituted f(x) by f'(x).

$u = x^2$

$du = 2xdx$

$\frac{1}{2}\int_{-2}^2 f(u) 2x dx \Rightarrow \frac{1}{2}\int_{-2}^2 f(u) du$

$\frac{1}{2}F(2) - \frac{1}{2}F(-2) \Rightarrow \frac{1}{2}F(2^2) - \frac{1}{2}F((-2)^2)$

I don't know the function $f$, but the definite integral ended up being $\frac{1}{2}F(4) - \frac{1}{2}F(4) = 0$.

 

[Mark44]

Oh, -_-'' I just noticed that I substituted f(x) by f'(x).

$u = x^2$

$du = 2xdx$

$\frac{1}{2}\int_{-2}^2 f(u) 2x dx \Rightarrow \frac{1}{2}\int_{-2}^2 f(u) du$

$\frac{1}{2}F(2) - \frac{1}{2}F(-2) \Rightarrow \frac{1}{2}F(2^2) - \frac{1}{2}F((-2)^2)$

I don't know the function $f$, but the definite integral ended up being $\frac{1}{2}F(4) - \frac{1}{2}F(4) = 0$.

That's what I get, as well, but your work is flaky here:
$\frac{1}{2}F(2) - \frac{1}{2}F(-2) \Rightarrow \frac{1}{2}F(2^2) - \frac{1}{2}F((-2)^2)$

Nit: Don't use $\Rightarrow$ in place of =. Expressions that have the same value should be connected with =. The arrow ($\Rightarrow$) should be used between statements (typically equations or inequalities) where the statement on the left implies the statement on the right.

Besides that, how do you justify changing F(2) to F(2²), and similar for F(-2) and F( (-2)²)?
The right way is to change the limits of integration to u values. When x = -2, u = ?, and when x = 2, u = ?.