a) Đúng vì trong tích phân \(\displaystyle \int\limits_0^1 {{x^n}{{(1 - x)}^m}dx} \), nếu đặt \(\displaystyle t = 1 - x\) thì \(\displaystyle dx = - dt\)
\(\displaystyle \Rightarrow \int\limits_0^1 {{x^n}{{(1 - x)}^m}dx} \) \(\displaystyle = \int\limits_1^0 {{{\left( {1 - t} \right)}^n}.{t^m}.\left( { - dt} \right)} \) \(\displaystyle = \int\limits_0^1 {{t^m}.{{\left( {1 - t} \right)}^n}dt} \) \(\displaystyle = \int\limits_0^1 {{x^m}.{{\left( {1 - x} \right)}^n}dt} \)
b) Ta có: \(\displaystyle \int\limits_{ - 1}^1 {\frac{{{t^2}dt}}{{{e^t} + 1}}} = \int\limits_{ - 1}^0 {\frac{{{t^2}dt}}{{{e^t} + 1}}} + \int\limits_0^1 {\frac{{{t^2}dt}}{{{e^t} + 1}}} \) (*)
Dùng phương pháp đổi biến \(\displaystyle t = - x\) đối với tích phân \(\displaystyle \int\limits_{ - 1}^0 {\frac{{{t^2}dt}}{{{e^t} + 1}}} \), ta được:
\(\displaystyle \int\limits_{ - 1}^0 {\frac{{{t^2}dt}}{{{e^t} + 1}}} = \int\limits_0^1 {\frac{{{x^2}dx}}{{{e^{ - x}} + 1}} = \int\limits_0^1 {\frac{{{t^2}dt}}{{{e^{ - t}} + 1}}} } \)\(\displaystyle = \int\limits_0^1 {\frac{{{t^2}{e^t}}}{{1 + {e^t}}}dt} \)
Thay vào (*) ta có: \(\displaystyle \int\limits_{ - 1}^1 {\frac{{{t^2}dt}}{{{e^t} + 1}}} \)\(\displaystyle = \int\limits_0^1 {\frac{{{t^2}{e^t}}}{{1 + {e^t}}}dt} + \int\limits_0^1 {\frac{{{t^2}dt}}{{{e^t} + 1}}} \) \(\displaystyle = \int\limits_0^1 {\frac{{{t^2}{e^t} + {t^2}}}{{{e^t} + 1}}dt} \) \(\displaystyle = \int\limits_0^1 {\frac{{{t^2}\left( {{e^t} + 1} \right)}}{{{e^t} + 1}}dt} = \int\limits_0^1 {{t^2}dt} \)
Vậy \(\displaystyle \int\limits_{ - 1}^1 {\frac{{{t^2}dt}}{{{e^t} + 1}}} = \int\limits_0^1 {{t^2}dt} \).
c) Sai.
Đặt \(\displaystyle \sin x = t \Rightarrow \cos xdx = dt\)
\(\displaystyle \Rightarrow \int\limits_0^1 {{{\sin }^3}s\cos xdx} \) \(\displaystyle = \int\limits_0^{\frac{\pi }{2}} {{t^3}dt} \ne \int\limits_0^1 {{t^3}dt} \)
Vậy c sai.
