Đề bài
Tính:
a) \(\int\limits_0^3 {{{\left( {3x - 1} \right)}^2}dx} \);
b) \(\int\limits_0^{\frac{\pi }{2}} {\left( {1 + \sin x} \right)dx} \);
c) \(\int\limits_0^1 {\left( {{e^{2x}} + 3{x^2}} \right)dx} \);
d) \(\int\limits_{ - 1}^2 {\left| {2x + 1} \right|dx} \).
Phương pháp giải - Xem chi tiết
Sử dụng kiến thức về tính chất của tích phân để tính: Cho f(x), g(x) là các hàm số liên tục trên đoạn [a; b]. Khi đó, ta có:
+ \(\int\limits_a^b {kf\left( x \right)dx} = k\int\limits_a^b {f\left( x \right)dx} \) (k là hằng số)
+ \(\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]dx} = \int\limits_a^b {f\left( x \right)dx} + \int\limits_a^b {g\left( x \right)dx} \)
+ \(\int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]dx} = \int\limits_a^b {f\left( x \right)dx} - \int\limits_a^b {g\left( x \right)dx} \)
+ \(\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^c {f\left( x \right)dx} + \int\limits_c^b {f\left( x \right)dx} \) \(\left( {a < c < b} \right)\)
Lời giải chi tiết
a) \(\int\limits_0^3 {{{\left( {3x - 1} \right)}^2}dx} = \int\limits_0^3 {\left( {9{x^2} - 6x + 1} \right)dx} = 9\int\limits_0^3 {{x^2}dx} - 6\int\limits_0^3 {xdx} + \int\limits_0^3 {dx} \)
\( = 3{x^3}\left| \begin{array}{l}3\\0\end{array} \right. - 3{x^2}\left| \begin{array}{l}3\\0\end{array} \right. + x\left| \begin{array}{l}3\\0\end{array} \right. = 81 - 27 + 3 = 57\)
b) \(\int\limits_0^{\frac{\pi }{2}} {\left( {1 + \sin x} \right)dx} = \int\limits_0^{\frac{\pi }{2}} {dx} + \int\limits_0^{\frac{\pi }{2}} {\sin xdx} = x\left| \begin{array}{l}\frac{\pi }{2}\\0\end{array} \right. - \cos x\left| \begin{array}{l}\frac{\pi }{2}\\0\end{array} \right. = \frac{\pi }{2} + 1\)
c) \(\int\limits_0^1 {\left( {{e^{2x}} + 3{x^2}} \right)dx} = \int\limits_0^1 {{{\left( {{e^2}} \right)}^x}dx} + 3\int\limits_0^1 {{x^2}dx} = \frac{{{e^{2x}}}}{{\ln {e^2}}}\left| \begin{array}{l}1\\0\end{array} \right. + {x^3}\left| \begin{array}{l}1\\0\end{array} \right. = \frac{{{e^2}}}{2} - \frac{1}{2} + 1 = \frac{{{e^2}}}{2} + \frac{1}{2}\)
d) \(\int\limits_{ - 1}^2 {\left| {2x + 1} \right|dx} = \int\limits_{ - 1}^{\frac{{ - 1}}{2}} {\left| {2x + 1} \right|dx} + \int\limits_{\frac{{ - 1}}{2}}^2 {\left| {2x + 1} \right|dx} = - \int\limits_{ - 1}^{\frac{{ - 1}}{2}} {\left( {2x + 1} \right)dx} + \int\limits_{\frac{{ - 1}}{2}}^2 {\left( {2x + 1} \right)dx} \)
\( = - \left( {{x^2} + x} \right)\left| \begin{array}{l}\frac{{ - 1}}{2}\\ - 1\end{array} \right. + \left( {{x^2} + x} \right)\left| \begin{array}{l}2\\\frac{{ - 1}}{2}\end{array} \right. = - \left[ {{{\left( {\frac{{ - 1}}{2}} \right)}^2} - \frac{1}{2} - {{\left( { - 1} \right)}^2} + 1} \right] + \left[ {{2^2} + 2 - {{\left( {\frac{{ - 1}}{2}} \right)}^2} + \frac{1}{2}} \right]\)
\( = \frac{1}{4} + \frac{{25}}{4} = \frac{{13}}{2}\)
