Lời giải

a) Vì \( \int_{-a}^{a} f(x) \, dx = \int_{-a}^{0} f(x) \, dx + \int_{0}^{a} f(x) \, dx \)

Xét \( \int_{0}^{a} f(x) \, dx \).

Đặt \( t = -x \implies x = -t \implies dx = -dt \)

\( \begin{cases}
x =0\\
x = a 
\end{cases}
\Rightarrow \begin{cases}
t = 0 \\
t = -a
\end{cases} \)

\( \int_{0}^{a} f(x) \, dx = \int_{0}^{-a} f(-t)(-dt) = -\int_{-a}^{0} f(-t) \, dt = -\int_{-a}^{0} f(x) \, dx \)

\( \Rightarrow \int_{-a}^{a} f(x) \, dx = 0 \)

b) Tương tự:

Ví dụ:
a) \( \int_{-\pi}^{\pi} \sin^3 x \, dx = 0 \) 

b) \( \int_{-1}^{1} (2x^3 - 4x) \, dx = 0\)

c) \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \ cos x \, dx = 0 \)  

d) \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \sin x \, dx = 0 \)