Giải bài tập 6 trang 27

Đề bài

Tính:

a) $\underset{0}{\overset{1}{\int }}(x^6-4x^3+3x^2)dx$

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

c) $\underset{1}{\overset{4}{\int }}\frac{1}{x\sqrt{x}}dx$

d) $\underset{0}{\overset{\frac{\pi }{2}}{\int }}(4\sin x+3\cos x)dx$

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

g) $\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{2}xdx$

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

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

k) $\underset{0}{\overset{1}{\int }}({3.4}^x-5e^{-x})dx$

Lời giải chi tiết

a) $\underset{0}{\overset{1}{\int }}(x^6-4x^3+3x^2)dx={\left(\frac{x^7}{7}-x^4+x^3\right)|}_0^1=\frac{1}{7}$

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

c) $\underset{1}{\overset{4}{\int }}\frac{1}{x\sqrt{x}}dx={\frac{-2}{\sqrt{x}}|}_1^4=1$

d) $\underset{0}{\overset{\frac{\pi }{2}}{\int }}(4\sin x+3\cos x)dx={\left(-4\cos x+3\sin x\right)|}_0^{\frac{\pi }{2}}=7$

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

g) $\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{2}xdx=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(\frac{1}{{\cos }^{2}x}-1\right)dx={\left(\tan x-x\right)|}_0^{\frac{\pi }{4}}=1-\frac{\pi }{4}$

h) $\underset{-1}{\overset{0}{\int }}{e}^{-x}dx=-{e^{-x}|}_{-1}^0=e-1$

i) $\underset{-2}{\overset{-1}{\int }}{e}^{x+2}dx={e^{x+2}|}_{-2}^{-1}=e-1$

k) $\underset{0}{\overset{1}{\int }}({3.4}^x-5e^{-x})dx={\left(3.\frac{4^x}{\ln 4}+5e^{-x}\right)|}_0^1=\frac{9}{\ln 4}+\frac{5}{e}-5$