Giải bài 3.16 trang 170 - SBT Đại số và Giải tích lớp 12

a)

\(\begin{aligned} & \int\limits_{0}^{1}{\left( {{y}^{3}}+3{{y}^{2}}-2 \right)dy}=\int\limits_{0}^{1}{{{y}^{3}}dy}+\int\limits_{0}^{1}{3{{y}^{2}}dy}-\int\limits_{0}^{1}{2dy} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{4}{{y}^{4}}\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right.+{{y}^{3}}\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right.-2y\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{4}+1-2\, \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-\frac{3}{4} \\ \end{aligned} \)

b) 

\(\begin{aligned} & \int\limits_{1}^{4}{\left( t+\frac{1}{\sqrt{t}}-\frac{1}{{{t}^{2}}} \right)dt}=\int\limits_{1}^{4}{tdt}+\int\limits_{1}^{4}{\frac{1}{\sqrt{t}}dt}-\int\limits_{1}^{4}{\frac{1}{{{t}^{2}}}dt} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{{{t}^{2}}}{2}\left| \begin{aligned} & 4 \\ & 1 \\ \end{aligned} \right.+2\sqrt{t}\left| \begin{aligned} & 4 \\ & 1 \\ \end{aligned} \right.+\frac{1}{t}\left| \begin{aligned} & 4 \\ & 1 \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \frac{16}{2}-\frac{1}{2} \right)+\left( 2\sqrt{4}-2 \right)+\left( \frac{1}{4}-1 \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{35}{4} \\ \end{aligned} \)

c)

\(\begin{aligned} & \int\limits_{0}^{\frac{\pi}{2}}{\left( 2\cos x -\sin 2x\right)dx}=\int\limits_{0}^{\frac{\pi}{2}}{2\cos xdx}-\int\limits_{0}^{\frac{\pi}{2}}{\sin 2xdx}\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sin x\left| \begin{aligned} & \frac{\pi}{2} \\ & 0 \\ \end{aligned} \right.+\dfrac{1}{2}\cos2x\left| \begin{aligned} & \frac{\pi}{2} \\ & 0 \\ \end{aligned} \right.\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2+\dfrac{1}{2}\left( -1-1 \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=1 \\ \end{aligned} \)

d) 

\(\begin{aligned} & \int\limits_{0}^{1}{{{\left( {{3}^{s}}-{{2}^{s}} \right)}^{2}}ds}=\int\limits_{0}^{1}{\left( {{9}^{s}}-{{2.6}^{s}}+{{4}^{s}} \right)ds} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int\limits_{0}^{1}{{{9}^{s}}ds}-2\int\limits_{0}^{1}{{{6}^{s}}ds}+\int\limits_{0}^{1}{{{4}^{s}}ds} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{{{9}^{s}}}{\ln 9}\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right.-2.\frac{{{6}^{s}}}{\ln 6}\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right.+\frac{{{4}^{s}}}{\ln 4}\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{8}{\ln 9}-\frac{10}{\ln 6}+\frac{3}{\ln 4} \\ \end{aligned} \)

e) 

\(\begin{aligned} & \int\limits_{0}^{\frac{\pi}{3}}{\cos 3xdx}+\int\limits_{\frac{\pi}{3}}^{\frac{3\pi}{2}}{\cos 3xdx}+\int\limits_{\frac{3\pi}{2}}^{\frac{5\pi}{2}}{\cos 3xdx}=\int\limits_{0}^{\frac{5\pi}{2}}{\cos 3xdx}\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{3}\sin3x\left| \begin{aligned} & \frac{5\pi}{2} \\ & 0 \\ \end{aligned} \right.\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{3}\left(\sin\dfrac{5\pi}{2}-\sin 0 \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-\dfrac{1}{3} \\ \end{aligned} \)

Ghi nhớ:

Giả sử F(x) là một nguyên hàm của hàm số f(x) trên đoạn \(\left[ a;b \right]\) . Khi đó

\(\int\limits_{a}^{b}{f\left( x \right)dx}=F\left( b \right)-F\left( a \right)\)