Đáp án:
\(\begin{array}{l}
29)\quad D.\, I = -5\\
30)\quad A.\, 1\\
31)\quad C.\, 7\\
33)\quad B.\, I = 2\displaystyle\int\limits_1^2(t^2 -t)dt
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
29)\quad \text{Ta có:}\\
\quad \displaystyle\int\limits_{-2}^4f(y)dy = \displaystyle\int\limits_{-2}^2f(y)dy+\displaystyle\int\limits_{2}^4f(y)dy\\
\to \displaystyle\int\limits_{2}^4f(y)dy=\displaystyle\int\limits_{-2}^4f(y)dy-\displaystyle\int\limits_{-2}^2f(y)dy\\
\to I = \displaystyle\int\limits_{-2}^4f(t)dt-\displaystyle\int\limits_{-2}^2f(x)dx\\
\to I = -4 - 1\\
\to I = -5\\
30)\quad I = \displaystyle\int\limits_{-1}^2[2f(x) + 3g(x)]dx\\
\to I = 2\displaystyle\int\limits_{-1}^2f(x)dx + 3\displaystyle\int\limits_{-1}^2g(x)dx\\
\to I = 2\cdot2 + 3\cdot (-1)\\
\to I = 1\\
31)\quad I = \displaystyle\int\limits_0^4f(x)dx\\
\to I = \displaystyle\int\limits_0^1f(x)dx+\displaystyle\int\limits_1^4f(x)dx\\
\to I = 2 + 5\\
\to I = 7\\
33)\quad I = \displaystyle\int\limits_0^3\dfrac{x}{1 + \sqrt{x+1}}dx\\
\to I = \displaystyle\int\limits_0^3(\sqrt{x+1} - 1)dx\\
\to I = \displaystyle\int\limits_0^32(\sqrt{x+1} - 1)\sqrt{x+1}\cdot \dfrac{1}{2\sqrt{x+1}}dx\\
Đặt\,\,t = \sqrt{x+1}\\
\to dt = \dfrac{1}{2\sqrt{x+1}}dx\\
\text{Đổi cận:}\\
x\quad \Big|\quad 0\qquad 3\\
\overline{t\,\quad \Big|\quad 1\qquad 2}\\
\text{Ta được:}\\
\quad I = \displaystyle\int\limits_1^22(t-1)tdt = 2\displaystyle\int\limits_1^2(t^2 -t)dt
\end{array}\)
