Giải bài 63 trang 30 sách bài tập toán 12 - Cánh diều
Tính: a) (intlimits_0^1 { - 2dx} ); b) (intlimits_0^1 {frac{{2x}}{3}dx} ); c) (intlimits_0^1 {{x^4}dx} ); d) (intlimits_1^3 {2sqrt[3]{x}dx} ); e) (intlimits_1^2 {frac{2}{{3x}}dx} ); g) (intlimits_1^9 {left( {xsqrt x - 2} right)dx} ).
Đề bài
Tính:
a) \(\int\limits_0^1 { - 2dx} \);
b) \(\int\limits_0^1 {\frac{{2x}}{3}dx} \);
c) \(\int\limits_0^1 {{x^4}dx} \);
d) \(\int\limits_1^3 {2\sqrt[3]{x}dx} \);
e) \(\int\limits_1^2 {\frac{2}{{3x}}dx} \); g) \(\int\limits_1^9 {\left( {x\sqrt x - 2} \right)dx} \).
Phương pháp giải - Xem chi tiết
Sử dụng các công thức:
• \(\int {{x^\alpha }dx} = \frac{{{x^{\alpha + 1}}}}{{\alpha + 1}} + C\).
• \(\int {\frac{1}{x}dx} = \ln \left| x \right| + C\).
Lời giải chi tiết
a) \(\int\limits_0^1 { - 2dx} = \left. { - 2{\rm{x}}} \right|_0^1 = - 2\).
b) \(\int\limits_0^1 {\frac{{2x}}{3}dx} = \left. {\frac{2}{3}{\rm{.}}\frac{{{{\rm{x}}^2}}}{2}} \right|_0^1 = \left. {\frac{{{{\rm{x}}^2}}}{3}} \right|_0^1 = \frac{1}{3}\).
c) \(\int\limits_0^1 {{x^4}dx} = \left. {\frac{{{x^5}}}{5}} \right|_0^1 = \frac{1}{5}\).
d) \(\int\limits_1^3 {2\sqrt[3]{x}dx} = \int\limits_1^3 {2{x^{\frac{1}{3}}}dx} = \left. {\frac{{2{{\rm{x}}^{\frac{4}{3}}}}}{{\frac{4}{3}}}} \right|_1^3 = \left. {\frac{{3x\sqrt[3]{x}}}{2}} \right|_1^3 = \frac{{9\sqrt[3]{3} - 3}}{2}\).
e) \(\int\limits_1^2 {\frac{2}{{3x}}dx} = \left. {\frac{2}{3}\ln \left| x \right|} \right|_1^2 = \frac{2}{3}\ln 2\).
g) \(\int\limits_1^9 {\left( {x\sqrt x - 2} \right)dx} = \int\limits_1^9 {\left( {{x^{\frac{3}{2}}} - 2} \right)dx} = \left. {\left( {\frac{{{{\rm{x}}^{\frac{5}{2}}}}}{{\frac{5}{2}}} - 2{\rm{x}}} \right)} \right|_1^9 = \left. {\left( {\frac{{2{{\rm{x}}^2}\sqrt x }}{5} - 2{\rm{x}}} \right)} \right|_1^9 = \frac{{396}}{5} - \left( { - \frac{8}{5}} \right) = \frac{{404}}{5}\).