第 12 类 RD Sharma 解 – 第 20 章定积分 – 练习 20.3 |设置 1

计算以下积分：

问题 1(i)。 $\int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}4x + 3 & , & \text{if }1 \leq x \leq 2 \\3x + 5 & , & \text{if }2 \leq x \leq 4\end{cases}$

解决方案：

We have,

I = $\int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}4x + 3 & , & \text{if }1 \leq x \leq 2 \\3x + 5 & , & \text{if }2 \leq x \leq 4\end{cases}$

I = $\int_1^4 f\left( x \right) dx$

Using additive property, we get

$I = \int_1^2 f\left( x \right) d x + \int_2^4 f\left( x \right) d x$

I = $\int_1^2 \left( 4x + 3 \right) dx + \int_2^4 \left( 3x + 5 \right) dx$

I = $\left[ 2 x^2 + 3x \right]_1^2 + \left[ \frac{3 x^2}{2} + 5x \right]_2^4$

I = 8 + 6 – 2 – 3 + 24 + 20 – 6 – 10

I = 37

问题 1(ii)。 $\int\limits_0^9 f\left( x \right) dx, where\ f\left( x \right)= \begin{cases}\sin x & , & 0 \leq x \leq \frac{\pi}{2} \\ 1 & , & \frac{\pi}{2} \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}$

解决方案：

We have,

I = $\int\limits_0^9 f\left( x \right) dx, where\ f\left( x \right)= \begin{cases}\sin x & , & 0 \leq x \leq \frac{\pi}{2} \\ 1 & , & \frac{\pi}{2} \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}$

I = $\int_0^9 f\left( x \right) d x$

Using additive property, we get

I = $\int_0^\frac{\pi}{2} f\left( x \right) d x + \int_\frac{\pi}{2}^3 f\left( x \right) d x + \int_3^9 f\left( x \right) d x$

I = $\int_0^\frac{\pi}{2} \sin x d x + \int_\frac{\pi}{2}^3 1 d x + \int_3^9 e^{x - 3} d x$

I = $\left[ - \cos x \right]_0^\frac{\pi}{2} + \left[ x \right]_\frac{\pi}{2}^3 + \left[ e^{x - 3} \right]_3^9$

I = 0 + 1 + 3 – π/2 + e⁶ – e⁰ 

I = 3 – π/2 + e⁶

问题 1(iii)。 $\int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}$

解决方案：

We have,

I = $\int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}$

I = $\int_1^4 f\left( x \right) d x$

Using additive property, we get

I = $\int_1^3 f\left( x \right) d x + \int_3^4 f\left( x \right) d x$

I = $\int_1^3 \left( 7x + 3 \right) d x + \int_3^4 8x d x$

I = $\left[ \frac{7 x^2}{2} + 3x \right]_1^3 + \left[ 4 x^2 \right]_3^4$

I = 63/2 + 9 – 7/2 – 3 + 64 – 36

I = 56/2 + 34

I = 62

问题2。 $\int\limits_{- 4}^4 \left| x + 2 \right| dx$

解决方案：

We have,

I = $\int\limits_{- 4}^4 \left| x + 2 \right| dx$

We know that, 

$\left| x + 2 \right| = \begin{cases} - \left( x + 2 \right) &, &- 4 \leq x \leq - 2 \\x + 2 &, &- 2 < x \leq 4\end{cases}$

So, we get

I = $\int_{- 4}^4 \left| x + 2 \right| d x$

I = $\int_{- 4}^{- 2} - \left( x + 2 \right) d x + \int_{- 2}^4 \left( x + 2 \right) d x$

I = $\left[ - \frac{x^2}{2} - 2x \right]_{- 4}^{- 2} + \left[ \frac{x^2}{2} + 2x \right]_{- 2}^4$

I = –2 + 4 – 8 – 8 + 8 + 8 – 2 + 4

I = 20

问题 3。 $\int\limits_{- 3}^3 \left| x + 1 \right| dx$

解决方案：

We have,

I = $\int_{- 3}^3 \left| x + 1 \right| d x$

We know that, 

$\left| x + 1 \right| = \begin{cases} - \left( x + 1 \right) &, &- 3 \leq x \leq - 1 \\x + 1 &, &- 1 < x \leq 3\end{cases}$

So, we get

I = $\int_{- 3}^{- 1} - \left( x + 1 \right) d x + \int_{- 1}^3 \left[ x + 1 \right] d x$

I = $\left[ - \frac{\left( x + 1 \right)^2}{2} \right]_{- 3}^{- 1} + \left[ \frac{\left( x + 1 \right)^2}{2} \right]_{- 1}^3$

I = 0 + 2 + 8 – 0

I = 10

问题 4。 $\int\limits_{- 1}^1 \left| 2x + 1 \right| dx$

解决方案：

We have,

I = $\int\limits_{- 1}^1 \left| 2x + 1 \right| dx$

We know that,

$\left| 2x + 1 \right| = \begin{cases} - \left( 2x + 1 \right) &, &- 1 \leq x \leq - \frac{1}{2} \\\left( 2x + 1 \right) &, &- \frac{1}{2} < x \leq 1\end{cases}$

So, we get

I = $\int_{- 1}^\frac{- 1}{2} - \left( 2x + 1 \right) d x + \int_{- \frac{1}{2}}^1 \left( 2x + 1 \right) d x$

I = $- \left[ x^2 + x \right]_{- 1}^\frac{- 1}{2} + \left[ x^2 + x \right]_{- \frac{1}{2}}^1$

I = –1/4 + 1/2 + 1 – 1 + 1 + 1 – 1/4 + 1/2

I = 5/2

问题 5。 $\int\limits_{- 2}^2 \left| 2x + 3 \right| dx$

解决方案：

We have,

I = $\int\limits_{- 2}^2 \left| 2x + 3 \right| dx$

We know that,

$\left| 2x + 3 \right| = \begin{cases} - \left( 2x + 3 \right) &, &- 2 \leq x \leq - \frac{3}{2}\\\left( 2x + 3 \right)&, &- \frac{3}{2} < x \leq 2\end{cases}$

So, we get

I = $\int_{- 2}^\frac{- 3}{2} - \left( 2x + 3 \right) d x + \int_{- \frac{3}{2}}^2 \left( 2x + 3 \right) d x$

I = $- \left[ x^2 + 3x \right]_{- 2}^\frac{- 3}{2} + \left[ x^2 + 3x \right]_{- \frac{3}{2}}^2$

I = –9/4 + 9/2 + 4 – 6 + 4 + 6 – 9/4 + 9/2

I = 25/2

问题 6。 $\int\limits_0^2 \left| x^2 - 3x + 2 \right| dx$

解决方案：

We have,

I = $\int\limits_0^2 \left| x^2 - 3x + 2 \right| dx$

We know that,

$\left| x^2 - 3x + 2 \right| = \begin{cases} - \left( x^2 - 3x + 2 \right)&, &\left( x - 1 \right)\left( x - 2 \right) \leq 0 \text{ or}, 1 \leq x \leq 2\\\left( x^2 - 3x + 2 \right)&, &x^2 - 3x + 2 \geq 0 \text{ or}, x \in \left( - \infty , 1 \right) \cup \left( 2, \infty \right)\end{cases}$

So we get,

I = $\int_0^2 \left( x^2 - 3x + 2 \right) d x$

I = $\int_0^1 \left( x^2 - 3x + 2 \right) d x - \int_1^2 \left( x^2 - 3x + 2 \right) d x$

I = $\left[ \frac{x^3}{3} - \frac{3 x^2}{2} + 2x \right]_0^1 - \left[ \frac{x^3}{3} - \frac{3 x^2}{2} + 2x \right]_1^2$

I = 1/3 – 3/2 + 2 – [8/3 – 6 + 4 – 1/3 + 3/2 – 2]

I = 1/3 – 3/2 + 2 – 8/3 + 6 – 2 + 1/3 – 3/2

I = 1

问题 7。 $\int\limits_0^3 \left| 3x - 1 \right| dx$

解决方案：

We have,

I = $\int\limits_0^3 \left| 3x - 1 \right| dx$

We know that,

$\left| 3x - 1 \right| = \begin{cases} - \left( 3x - 1 \right)&,&0 \leq x \leq \frac{1}{3}\\\left( 3x - 1 \right)&,& \frac{1}{3} < x \leq 3\end{cases}$

So we get,

I = $\int_0^\frac{1}{3} - \left( 3x + 1 \right) dx + \int_\frac{1}{3}^0 \left( 3x + 1 \right) dx$

I = $\left[ \frac{- 3 x^2}{2} - x \right]_0^\frac{1}{3} + \left[ \frac{3 x^2}{2} + x \right]_\frac{1}{3}^3$

I = –1/6 + 1/3 – 0 + 27/2 + 3 – 1/6 – 1/3

I = 65/6

问题 8。 $\int\limits_{- 6}^6 \left| x + 2 \right| dx$

解决方案：

We have,

I = $\int\limits_{- 6}^6 \left| x + 2 \right| dx$

We know that,

$\left| x + 2 \right| = \begin{cases} - \left( x + 2 \right) &,& - 6 \leq x \leq - 2\\x + 2&,& - 2 < x \leq 6\end{cases}$

So, we get 

I = $\int_{- 6}^6 \left| x + 2 \right| d x$

I = $\int_{- 6}^{- 2} - \left( x + 2 \right) dx + \int_{- 2}^6 \left( x + 2 \right) dx$

I = $\left[ \frac{- x^2}{2} - 2x \right]_{- 6}^{- 2} + \left[ \frac{x^2}{2} + 2x \right]_{- 2}^6$

I = –2 + 4 + 18 – 12 + 18 + 12 – 2 + 4

I = 40

问题 9。 $\int\limits_{- 2}^2 \left| x + 1 \right| dx$

解决方案：

We have,

I = $\int\limits_{- 2}^2 \left| x + 1 \right| dx$

We know that,

$\left| x + 1 \right| = \begin{cases} - \left( x + 1 \right) &,& - 2 \leq x \leq - 1\\x + 1&,& - 1 < x \leq 2\end{cases}$

So we get,

I = $\int_{- 2}^2 \left| x + 1 \right| d x$

I = $\int_{- 2}^{- 1} - \left( x + 1 \right) dx + \int_{- 1}^2 \left( x + 1 \right) dx$

I = $\left[ \frac{- x^2}{2} - x \right]_{- 2}^{- 1} + \left[ \frac{x^2}{2} + x \right]_{- 1}^2$

I = -1/2 + 1 + 2 – 2 + 2 + 2 – 1/2 + 1

I = 5

问题 10。 $\int\limits_1^2 \left| x - 3 \right| dx$

解决方案：

We have,

I = $\int\limits_1^2 \left| x - 3 \right| dx$

We know that,

$\left| x + 1 \right| = \begin{cases} - \left( x + 1 \right) &,& 1 \leq x \leq 3\\\left( x + 1 \right)&,& x > 3\end{cases}$

So we get,

I = $\int_1^2 \left| x - 3 \right| d x$

I = $\int_1^2 - \left( x - 3 \right) dx$

I = $\left[ \frac{- x^2}{2} - 3x \right]_1^2$

I = – 2 – 6 + 1/2 + 3

I = – 5 + 1/2

I = (-10 + 1)/2

I = -9/2

问题 11。 $\int\limits_0^{\pi/2} \left| \cos 2x \right| dx$

解决方案：

We have,

I = $\int\limits_0^{\pi/2} \left| \cos 2x \right| dx$

We know that,

$\left| \cos 2x \right| = \begin{cases} - \cos 2x &,& \frac{\pi}{4} \leq x \leq \frac{\pi}{2}\\\cos 2x&,& 0 < x \leq \frac{\pi}{4}\end{cases}$

So we get,

I = $\int_{- 2}^2 \left| \cos 2x \right| d x$

I = $\int_0^\frac{\pi}{4} \cos 2x dx - \int_\frac{\pi}{4}^\frac{\pi}{2} \cos 2x dx$

I = $\left[ \frac{\sin 2x}{2} \right]_0^\frac{\pi}{4} - \left[ \frac{\sin 2x}{2} \right]_\frac{\pi}{4}^\frac{\pi}{2}$

I = 1/2 – 0 – 0 + 1/2

I = 1

问题 12。 $\int\limits_0^{2\pi} \left| \sin x \right| dx$

解决方案：

We have,

I = $\int\limits_0^{2\pi} \left| \sin x \right| dx$

We know that,

$\left| \sin x \right| = \begin{cases} - \sin x &,& \pi \leq x \leq 2\pi\\\sin x&,& 0 < x \leq \pi\end{cases}$

So we get,

I = $\int_0^{2\pi} \left| \sin x \right| dx$

I = $\int_0^\pi \sin x dx + \int_\pi^{2\pi} - \sin x dx$

I = $- \left[ \cos x \right]_0^\pi + \left[ \cos x \right]_\pi^{2\pi}$

I = 1 + 1 + 1 – (–1)

I = 1 + 1 + 1 + 1

I = 4

问题 13。 $\int\limits_{- \pi/4}^{\pi/4} \left| \sin x \right| dx$

解决方案：

We have,

I = $\int\limits_{- \pi/4}^{\pi/4} \left| \sin x \right| dx$

We know that,

$\left| \sin x \right| = \begin{cases} - \sin x &,& - \frac{\pi}{4} \leq x \leq 0\\\sin x&,& 0 < x \leq \frac{\pi}{4}\end{cases}$

So we get,

I = $\int_{- \frac{\pi}{4}}^\frac{\pi}{4} \left| \sin x \right| d x$

I = $\int_{- \frac{\pi}{4}}^0 - \sin x dx + \int_0^\frac{\pi}{4} \sin x dx$

I = $\left[ \cos x \right]_\frac{- \pi}{4}^0 - \left[ \cos x \right]_0^\frac{- \pi}{4}$

I = 1 – 1/√2 – 1/√2 + 1

I = 2 – 2/√2

I = 2 – √2

问题 14。 $\int\limits_2^8 \left| x - 5 \right| dx$

解决方案：

We have,

I = $\int\limits_2^8 \left| x - 5 \right| dx$

We know that,

$\left| x - 5 \right| = \begin{cases} - \left( x - 5 \right) &,& 2 \leq x \leq 5\\x - 5&,& 5 < x \leq 8\end{cases}$

So we get,

I = $\int_2^8 \left| x - 5 \right| d x$

I = $\int_2^5 - \left( x - 5 \right) dx + \int_5^8 \left( x - 5 \right) dx$

I = $- \left[ \frac{x^2}{2} - 5x \right]_2^5 + \left[ \frac{x^2}{2} - 5x \right]_5^8$

I = – 25/2 + 25 + 2 – 10 + 32 – 40 – 25/2 + 25

I = 9