a) Khai triển bình phương:
\[
\int\limits_0^3 (3x-1)^2\,dx = \int\limits_0^3 (9x^2 - 6x + 1)\,dx
\]
\[
= 9\int\limits_0^3 x^2\,dx - 6\int\limits_0^3 x\,dx + \int\limits_0^3 dx
\]
\[
= \left[3x^3\right]_0^3 - \left[3x^2\right]_0^3 + \left[x\right]_0^3
\]
\[
= 81 - 27 + 3 = 57.
\]
b) Tách tích phân:
\[
\int\limits_0^{\frac{\pi}{2}} (1 + \sin x)\,dx = \int\limits_0^{\frac{\pi}{2}} dx + \int\limits_0^{\frac{\pi}{2}} \sin x\,dx
\]
\[
= \left[x\right]_0^{\frac{\pi}{2}} + \left[-\cos x\right]_0^{\frac{\pi}{2}}
\]
\[
= \frac{\pi}{2} + (-\cos\tfrac{\pi}{2} + \cos 0) = \frac{\pi}{2} + (0 + 1) = \frac{\pi}{2} + 1.
\]
c) Tách tích phân và dùng nguyên hàm \(\int e^{2x}dx = \dfrac{e^{2x}}{2}\):
\[
\int\limits_0^1 (e^{2x} + 3x^2)\,dx = \int\limits_0^1 e^{2x}\,dx + 3\int\limits_0^1 x^2\,dx
\]
\[
= \left[\frac{e^{2x}}{2}\right]_0^1 + \left[x^3\right]_0^1
\]
\[
= \frac{e^2}{2} - \frac{1}{2} + 1 = \frac{e^2}{2} + \frac{1}{2}.
\]
d) Tìm điểm \(2x+1=0 \Rightarrow x = -\dfrac{1}{2}\). Điểm này nằm trong \((-1;\,2)\) nên chia đoạn:
\[
\int\limits_{-1}^2 |2x+1|\,dx = \int\limits_{-1}^{-\frac{1}{2}} |2x+1|\,dx + \int\limits_{-\frac{1}{2}}^2 |2x+1|\,dx.
\]
Trên \(\left[-1;\,-\dfrac{1}{2}\right]\): \(2x+1 \leq 0\) nên \(|2x+1| = -(2x+1)\).
Trên \(\left[-\dfrac{1}{2};\,2\right]\): \(2x+1 \geq 0\) nên \(|2x+1| = 2x+1\).
Nguyên hàm của \(2x+1\) là \(x^2+x\). Do đó:
\[
= -\left[x^2+x\right]_{-1}^{-\frac{1}{2}} + \left[x^2+x\right]_{-\frac{1}{2}}^2
\]
\[
= -\left[\left(\frac{1}{4}-\frac{1}{2}\right) - (1-1)\right] + \left[(4+2) - \left(\frac{1}{4}-\frac{1}{2}\right)\right]
\]
\[
= -\left(-\frac{1}{4}\right) + \left(6+\frac{1}{4}\right) = \frac{1}{4} + \frac{25}{4} = \frac{26}{4} = \frac{13}{2}.
\]
