Giải bài tập 6 trang 27 SGK Toán 12 tập 2 - Cánh diều — Không quảng cáo

Đề bài

Tính:

a) $\int\limits_0^1 {({x^6} - 4{x^3} + 3{x^2})dx}$

b) $\int\limits_1^2 {\frac{1}{{{x^4}}}dx}$

c) $\int\limits_1^4 {\frac{1}{{x\sqrt x }}dx}$

d) $\int\limits_0^{\frac{\pi }{2}} {(4\sin x + 3\cos x)dx}$

e) $\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {{{\cot }^2}xdx}$

g) $\int\limits_0^{\frac{\pi }{4}} {{{\tan }^2}xdx}$

h) $\int\limits_{ - 1}^0 {{e^{ - x}}dx}$

i) $\int\limits_{ - 2}^{ - 1} {{e^{x + 2}}dx}$

k) $\int\limits_0^1 {({{3.4}^x} - 5{e^{ - x}})dx}$

Lời giải chi tiết

a) $\int\limits_0^1 {({x^6} - 4{x^3} + 3{x^2})dx}  = \left. {\left( {\frac{{{x^7}}}{7} - {x^4} + {x^3}} \right)} \right|_0^1 = \frac{1}{7}$

b) $\int\limits_1^2 {\frac{1}{{{x^4}}}dx}  = \left. {\left( { - \frac{1}{{3{x^3}}}} \right)} \right|_1^2 = \frac{7}{{24}}$

c) $\int\limits_1^4 {\frac{1}{{x\sqrt x }}dx}  = \left. {\frac{{ - 2}}{{\sqrt x }}} \right|_1^4 = 1$

d) $\int\limits_0^{\frac{\pi }{2}} {(4\sin x + 3\cos x)dx}  = \left. {\left( { - 4\cos x + 3\sin x} \right)} \right|_0^{\frac{\pi }{2}} = 7$

e) $\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {{{\cot }^2}xdx}  = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\left( {\frac{1}{{{{\sin }^2}x}} - 1} \right)dx}  = \left. {\left( { - \cot x - x} \right)} \right|_{\frac{\pi }{4}}^{\frac{\pi }{2}} =  - \frac{\pi }{2} - ( - 1 - \frac{\pi }{4}) = 1 - \frac{\pi }{4}$

g) $\int\limits_0^{\frac{\pi }{4}} {{{\tan }^2}xdx}  = \int\limits_0^{\frac{\pi }{4}} {\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)dx}  = \left. {\left( {\tan x - x} \right)} \right|_0^{\frac{\pi }{4}} = 1 - \frac{\pi }{4}$

h) $\int\limits_{ - 1}^0 {{e^{ - x}}dx}  =  - \left. {{e^{ - x}}} \right|_{ - 1}^0 = e - 1$

i) $\int\limits_{ - 2}^{ - 1} {{e^{x + 2}}dx}  = \left. {{e^{x + 2}}} \right|_{ - 2}^{ - 1} = e - 1$

k) $\int\limits_0^1 {({{3.4}^x} - 5{e^{ - x}})dx}  = \left. {\left( {3.\frac{{{4^x}}}{{\ln 4}} + 5{e^{ - x}}} \right)} \right|_0^1 = \frac{9}{{\ln 4}} + \frac{5}{e} - 5$