第 12 类 RD Sharma 解决方案 – 第 20 章定积分 – 练习 20.5 |设置 2

将以下定积分计算为和的极限：

问题 12。 $\int_{0}^{2}(x^2+4)dx$

解决方案：

We have,

I =$\int_{0}^{2}(x^2+4)dx$

We know,

$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$, where h =$\frac{b-a}{n}$

Here a = 0, b = 2 and f(x) = x² + 4.

=> h = 2/n

=> nh = 2

So, we get,

I =$\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$

=$\lim_{h\to0}h[4+(h^2+4)+((2h)^2+4)+...+(((n-1)h)^2+4)]$

=$\lim_{h\to0}h[4n+h^2(1^2+2^2+3^2+...+(n-1)^2)]$

=$\lim_{h\to0}h[4n+h^2(1^2+2^2+3^2+...+(n-1)^2)]$

=$\lim_{h\to0}h[4n+h^2(\frac{n(n-1)(2n-1)}{6})]$

Now if h −> 0, then n −> ∞. So, we have,

=$\lim_{n\to\infty}\frac{2}{n}[4n+\frac{4}{n^2}(\frac{n(n-1)(2n-1)}{6})]$

=$\lim_{n\to\infty}[8+\frac{4n^3}{3n^3}(1-\frac{1}{n})(1-\frac{2}{n})]$

= 8 +$\frac{4(2)}{3}$

= 8 +$\frac{8}{3}$

=$\frac{32}{3}$

Therefore, the value of$\int_{0}^{2}(x^2+4)dx$as limit of sum is$\frac{32}{3}$.

问题 13。 $\int_{1}^{4}(x^2-x)dx$

解决方案：

We have,

I =$\int_{1}^{4}(x^2-x)dx$

We know,

$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$, where h =$\frac{b-a}{n}$

Here a = 1, b = 4 and f(x) = x² − x.

=> h = 3/n

=> nh = 3

So, we get,

I =$\lim_{h\to0}h[f(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]$

=$\lim_{h\to0}h[(1^2-1)+[(1+h)^2-(1+h)]+[(1+2h)^2-(1+2h)]+...+[(1+(n-1)h)^2-(1+(n-1)h)]]$

=$\lim_{h\to0}h[0+[h+h^2]+[2h+(2h)^2]+...+[(n-1)h+((n-1)h)^2]]$

=$\lim_{h\to0}h[h(1+2+3+...+(n-1))+h^2(1^2+2^2+3^2+...+(n-1)^2)]$

=$\lim_{h\to0}h[h(\frac{n(n-1)}{2})+h^2(\frac{n(n-1)(2n-1)}{6})]$

Now if h −> 0, then n −> ∞. So, we have,

=$\lim_{n\to\infty}\frac{3}{n}[\frac{3}{n}(\frac{n(n-1)}{2})+\frac{9}{n^2}(\frac{n(n-1)(2n-1)}{6})]$

=$\lim_{n\to\infty}[\frac{9n^2}{2n^2}(1-\frac{1}{n})+\frac{3n^3}{2n^3}(1-\frac{1}{n})(1-\frac{2}{n})]$

=$\frac{9}{2}+3$

=$\frac{15}{2}$

Therefore, the value of$\int_{1}^{4}(x^2-x)dx$as limit of sum is$\frac{15}{2}$.

问题 14。 $\int_{0}^{1}(3x^2+5x)dx$

解决方案：

We have,

I =$\int_{0}^{1}(3x^2+5x)dx$

We know,

$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$, where h =$\frac{b-a}{n}$

Here a = 0, b = 1 and f(x) = 3x² + 5x.

=> h = 1/n

=> nh = 1

So, we get,

I =$\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$

=$\lim_{h\to0}h[0+[3h^2+5h]+[3(2h)^2+5(2h)]+...+[3((n-1)h)^2+5((n-1)h)]$

=$\lim_{h\to0}h[3h^2(1^2+2^2+3^2+...+(n-1)^2)+5h(1+2+3+...+(n-1))]$

=$\lim_{h\to0}h[3h^2(\frac{n(n-1)(2n-1)}{6})+5h(\frac{n(n-1)}{2})]$

Now if h −> 0, then n −> ∞. So, we have,

=$\lim_{n\to\infty}\frac{1}{n}[\frac{3}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{5}{n}(\frac{n(n-1)}{2})]$

=$\lim_{n\to\infty}[\frac{n^3}{2n^3}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{5n^2}{2n^2}(1-\frac{1}{n})]$

= 1 +$\frac{5}{2}$

=$\frac{7}{2}$

Therefore, the value of$\int_{0}^{1}(3x^2+5x)dx$as limit of sum is$\frac{7}{2}$.

问题 15。 $\int_{0}^{2}e^xdx$

解决方案：

We have,

I =$\int_{0}^{2}e^xdx$

We know,

$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$, where h =$\frac{b-a}{n}$

Here a = 0, b = 2 and f(x) = e^x.

=> h = 2/n

=> nh = 2

So, we get,

I =$\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$

=$\lim_{h\to0}h[1+e^h+e^{2h}+...+e^{(n-1)h}]$

=$\lim_{h\to0}h[\frac{(e^h)^n-1}{e^h-1}]$

=$\lim_{h\to0}h[\frac{e^{nh}-1}{e^h-1}]$

=$\lim_{h\to0}h[\frac{e^{2}-1}{e^h-1}]$

=$\lim_{h\to0}\left(\frac{e^{2}-1}{\frac{e^h-1}{h}}\right)$

= e² − 1

Therefore, the value of$\int_{0}^{2}e^xdx$as limit of sum is e² − 1.

问题 16。 $\int_{a}^{b}e^xdx$

解决方案：

We have,

I =$\int_{a}^{b}e^xdx$

We know,

$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$, where h =$\frac{b-a}{n}$

Here a = a, b = b and f(x) = e^x.

=> h =$\frac{b-a}{n}$

=> nh = b − a

So, we get,

I =$\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$

=$\lim_{h\to0}h[e^a+e^{a+h}+e^{a+2h}+...+e^{a+(n-1)h}]$

=$\lim_{h\to0}he^a[1+e^{h}+e^{2h}+...+e^{(n-1)h}]$

=$\lim_{h\to0}he^a[\frac{(e^h)^n-1}{e^h-1}]$

=$\lim_{h\to0}he^a[\frac{e^{nh}-1}{e^h-1}]$

=$\lim_{h\to0}he^a[\frac{e^{b-a}-1}{e^h-1}]$

=$\lim_{h\to0}e^a\left(\frac{e^{b-a}-1}{\frac{e^h-1}{h}}\right)$

= e^a (e^b-a −1)

= e^b − e^a

Therefore, the value of$\int_{a}^{b}e^xdx$as limit of sum is e^b − e^a.

问题 17。 $\int_{a}^{b}cosxdx$

解决方案：

We have,

I =$\int_{a}^{b}cosxdx$

We know,

$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$, where h =$\frac{b-a}{n}$

Here a = a, b = b and f(x) = cos x.

=> h =$\frac{b-a}{n}$

=> nh = b − a

So, we get,

I =$\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$

=$\lim_{h\to0}h[cosa+cos(a+h)+cos(a+2h)+...+cos(a+(n-1)h)]$

=$\lim_{h\to0}h\left[\frac{cos(a+(n-1)\frac{h}{2})sin\frac{nh}{2}}{sin\frac{h}{2}}\right]$

=$\lim_{h\to0}h\left[\frac{cos(a+\frac{nh}{2}-\frac{h}{2})sin\frac{nh}{2}}{sin\frac{h}{2}}\right]$

=$\lim_{h\to0}h\left[\frac{cos(a+\frac{b-a}{2}-\frac{h}{2})sin\frac{b-a}{2}}{sin\frac{h}{2}}\right]$

=$\lim_{h\to0}\left[\frac{\frac{h}{2}}{sin\frac{h}{2}}×2cos(a+\frac{b-a}{2}-\frac{h}{2})(sin\frac{b-a}{2})\right]$

=$\lim_{h\to0}\left[2cos(a+\frac{b-a}{2})sin(\frac{b-a}{2})\right]$

=$2cos(\frac{a+b}{2})sin(\frac{b-a}{2})$

= sin b − sin a

Therefore, the value of$\int_{a}^{b}cosxdx$as limit of sum is sin b − sin a.

问题 18。 $\int_{0}^{\frac{\pi}{2}}sinxdx$

解决方案：

We have,

I =$\int_{0}^{\frac{\pi}{2}}sinxdx$

We know,

$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$, where h =$\frac{b-a}{n}$

Here a = 0, b =$\frac{\pi}{2}$and f(x) = sin x.

=> h =$\frac{\pi}{2n}$

=> nh =$\frac{2}{\pi}$

So, we get,

I =$\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$

=$\lim_{h\to0}h[sin0+sinh+sin2h+...+sin(n-1)h]$

=$\lim_{h\to0}h\frac{sin(\frac{(n-1)h}{2})sin(\frac{nh}{2})}{sin\frac{h}{2}}$

=$\lim_{h\to0}h\frac{sin(\frac{nh}{2}-\frac{h}{2})sin(\frac{nh}{2})}{sin\frac{h}{2}}$

=$\lim_{h\to0}h\frac{sin(\frac{\pi}{4}-\frac{h}{2})sin(\frac{\pi}{4})}{sin\frac{h}{2}}$

=$2(\frac{1}{2})$

= 1

Therefore, the value of$\int_{0}^{\frac{\pi}{2}}sinxdx$as limit of sum is 1.

问题 19。 $\int_{0}^{\frac{\pi}{2}}cosxdx$

解决方案：

We have,

I =$\int_{0}^{\frac{\pi}{2}}cosxdx$

We know,

$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$, where h =$\frac{b-a}{n}$

Here a = 0, b =$\frac{\pi}{2}$and f(x) = cos x.

=> h =$\frac{\pi}{2n}$

=> nh =$\frac{2}{\pi}$

So, we get,

I =$\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$

=$\lim_{h\to0}h[cos0+cosh+cos2h+...+cos(n-1)h]$

=$\lim_{h\to0}h[\frac{cos(\frac{nh}{2}-\frac{h}{2})cos\frac{nh}{2}}{cos\frac{h}{2}}]$

=$\lim_{h\to0}h[\frac{cos(\frac{\pi}{4}-\frac{h}{2})cos\frac{\pi}{4}}{cos\frac{h}{2}}]$

=$2(\frac{1}{2})$

= 1

Therefore, the value of$\int_{0}^{\frac{\pi}{2}}cosxdx$as limit of sum is 1.

问题 20。 $\int_{1}^{4}(3x^2+2x)dx$

解决方案：

We have,

I =$\int_{1}^{4}(3x^2+2x)dx$

We know,

$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$, where h =$\frac{b-a}{n}$

Here a =1, b = 4 and f(x) = 3x² + 2x.

=> h = 3/n

=> nh = 3

So, we get,

I =$\lim_{h\to0}h[f(0)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]$

=$\lim_{h\to0}h[(3+2)+[3(1+h)^2+2(1+h)]+[3(1+2h)^2+2(1+2h)]+...+[3(1+(n-1)h)^2+2(1+(n-1)h)]]$

=$\lim_{h\to0}h[5n+8h(1+2+3...+(n-1))+3h^2(1^2+2^2+3^2+...+(n-1)^2)]$

=$\lim_{h\to0}h[5n+8h(\frac{n(n-1)}{2})+3h^2(\frac{n(n-1)(2n-1)}{6})]$

Now if h −> 0, then n −> ∞. So, we have,

=$\lim_{n\to\infty}\frac{3}{n}[5n+\frac{24}{n}(\frac{n(n-1)}{2})+\frac{27}{n^2}(\frac{n(n-1)(2n-1)}{6})]$

=$\lim_{n\to\infty}[15+\frac{36n^2}{n^2}(1-\frac{1}{n})+\frac{27n^3}{2n^3}(1-\frac{1}{n})(2-\frac{1}{n})]$

= 15 + 36 + 27

= 78

Therefore, the value of$\int_{1}^{4}(3x^2+2x)dx$as limit of sum is 78.

问题 21。 $\int_{0}^{2}(3x^2-2)dx$

解决方案：

We have,

I =$\int_{0}^{2}(3x^2-2)dx$

We know,

$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$, where h =$\frac{b-a}{n}$

Here a =0, b = 2 and f(x) = 3x² − 2.

=> h = 2/n

=> nh = 2

So, we get,

I =$\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$

=$\lim_{h\to0}h[-2+[3h^2-2]+[3(2h)^2-2]+...+[3((n-1)h)^2-2]]$

=$\lim_{h\to0}h[-2n+3h^2(1^2+2^2+3^2+...+(n-1)^2)]$

=$\lim_{h\to0}h[-2n+3h^2(\frac{n(n-1)(2n-1)}{6})]$

Now if h −> 0, then n −> ∞. So, we have,

=$\lim_{n\to\infty}\frac{2}{n}[-2n+\frac{12}{n^2}(\frac{n(n-1)(2n-1)}{6})]$

=$\lim_{n\to\infty}[-4+\frac{4n^3}{n^3}(1-\frac{1}{n})(1-\frac{2}{n})]$

= −4 + 8

= 4

Therefore, the value of$\int_{0}^{2}(3x^2-2)dx$as limit of sum is 4.

问题 22。 $\int_{0}^{2}(x^2+2)dx$

解决方案：

We have,

I =$\int_{0}^{2}(x^2+2)dx$

We know,

$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$, where h =$\frac{b-a}{n}$

Here a =0, b = 2 and f(x) = x² + 2.

=> h = 2/n

=> nh = 2

So, we get,

I =$\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$

=$\lim_{h\to0}h[2+(h^2+2)+[((2h)^2+2)]+...+[((n-1)h)^2+2]]$

=$\lim_{h\to0}h[2n+h^2(1^2+2^2+3^2+...+(n-1)^2)]$

=$\lim_{h\to0}h[2n+h^2(\frac{n(n-1)(2n-1)}{6})]$

Now if h −> 0, then n −> ∞. So, we have,

=$\lim_{n\to\infty}\frac{2}{n}[2+\frac{4}{n^2}(\frac{n(n-1)(2n-1)}{6})]$

=$\lim_{n\to\infty}[4+\frac{4n^3}{n^3}(1-\frac{1}{n})(1-\frac{2}{n})]$

= 4 +$\frac{4(2)}{3}$

= 4 +$\frac{8}{3}$

=$\frac{20}{3}$

Therefore, the value of$\int_{0}^{2}(x^2+2)dx$as limit of sum is$\frac{20}{3}$.