To show that the functions [tex]\( f_1(x) = e^x \)[/tex] and [tex]\( f_2(x) = x e^{-x} - e^{-x} \)[/tex] are orthogonal on the interval [tex]\([0, 2]\)[/tex], we need to compute the inner product of these two functions over the given interval and show that this inner product is zero.
The inner product of two functions [tex]\( f_1(x) \)[/tex] and [tex]\( f_2(x) \)[/tex] over the interval [tex]\([a, b]\)[/tex] is defined as:
[tex]\[
\langle f_1, f_2 \rangle = \int_a^b f_1(x) f_2(x) \, dx
\][/tex]
In this case, we need to compute:
[tex]\[
\langle f_1, f_2 \rangle = \int_0^2 e^x \left( x e^{-x} - e^{-x} \right) \, dx
\][/tex]
Let's simplify the integrand first:
[tex]\[
e^x \left( x e^{-x} - e^{-x} \right) = e^x x e^{-x} - e^x e^{-x}
\][/tex]
Then:
[tex]\[
e^x x e^{-x} = x
\][/tex]
and
[tex]\[
e^x e^{-x} = 1
\][/tex]
Thus, the integrand becomes:
[tex]\[
x - 1
\][/tex]
Now, the integral to compute is:
[tex]\[
\int_0^2 (x - 1) \, dx
\][/tex]
We can split this integral into two simpler integrals:
[tex]\[
\int_0^2 (x - 1) \, dx = \int_0^2 x \, dx - \int_0^2 1 \, dx
\][/tex]
First, compute [tex]\(\int_0^2 x \, dx\)[/tex]:
[tex]\[
\int_0^2 x \, dx = \left[ \frac{x^2}{2} \right]_0^2 = \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} = 2
\][/tex]
Next, compute [tex]\(\int_0^2 1 \, dx\)[/tex]:
[tex]\[
\int_0^2 1 \, dx = \left[ x \right]_0^2 = 2 - 0 = 2
\][/tex]
Putting it all together:
[tex]\[
\int_0^2 (x - 1) \, dx = 2 - 2 = 0
\][/tex]
Since the inner product [tex]\(\langle f_1, f_2 \rangle\)[/tex] is zero, this shows that the functions [tex]\( f_1(x) = e^x \)[/tex] and [tex]\( f_2(x) = x e^{-x} - e^{-x} \)[/tex] are indeed orthogonal on the interval [tex]\([0, 2]\)[/tex].
