Question

Tính và so sánh:
a) $\int_0^1 2 x d x$ và $2 \int_0^1 x d x$;
b) $\int_0^1\left(x^2+x\right) d x$ và $\int_0^1 x^2 d x+\int_0^1 x d x$;
c) $\int_0^3 x d x$ và $\int_0^1 x d x+\int_1^3 x d x$.

Answer 1

a) Ta có: \(\int\limits_0^1 {2xdx}  = {x^2}\left| \begin{array}{l}1\\0\end{array} \right. = 1\), \(2\int\limits_0^1 {xdx}  = 2.\frac{{{x^2}}}{2}\left| \begin{array}{l}1\\0\end{array} \right. = 1\) nên \(\int\limits_0^1 {2xdx}  = 2\int\limits_0^1 {xdx} \)

b) Ta có: \(\int\limits_0^1 {\left( {{x^2} + x} \right)dx}  = \left( {\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2}} \right)\left| \begin{array}{l}1\\0\end{array} \right. = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}\)

\(\int\limits_0^1 {{x^2}dx}  + \int\limits_0^1 {xdx}  = \frac{{{x^3}}}{3}\left| \begin{array}{l}1\\0\end{array} \right. + \frac{{{x^2}}}{2}\left| \begin{array}{l}1\\0\end{array} \right. = \frac{1}{3} - 0 + \frac{1}{2} - 0 = \frac{5}{6}\)

Do đó, \(\int\limits_0^1 {\left( {{x^2} + x} \right)dx}  = \int\limits_0^1 {{x^2}dx}  + \int\limits_0^1 {xdx} \)

c) Ta có: \(\int\limits_0^3 {xdx}  = \frac{{{x^2}}}{2}\left| \begin{array}{l}3\\0\end{array} \right. = \frac{{{3^2}}}{2} - 0 = \frac{9}{2}\); \(\int\limits_0^1 {xdx}  + \int\limits_1^3 {xdx}  = \frac{{{x^2}}}{2}\left| \begin{array}{l}1\\0\end{array} \right. + \frac{{{x^2}}}{2}\left| \begin{array}{l}3\\1\end{array} \right. = \frac{1}{2} - 0 + \frac{{{3^2}}}{2} - \frac{1}{2} = \frac{9}{2}\)

Do đó, \(\int\limits_0^3 {xdx}  = \int\limits_0^1 {xdx}  + \int\limits_1^3 {xdx} \)