7.3E: Mazoezi ya Sehemu ya 7.3

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Edwin “Jed” Herman & Gilbert Strang
OpenStax

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Punguza maneno katika mazoezi 1 - 5 kwa kuandika kila mmoja kwa kutumia kazi moja ya trigonometric.

1)\(4−4\sin^2θ\)

2)\(9\sec^2θ−9\)

Jibu: \(9\sec^2θ−9 \quad = \quad 9\tan^2θ\)

3)\(a^2+a^2\tan^2θ\)

4)\(a^2+a^2\sinh^2θ\)

Jibu: \(a^2+a^2\sinh^2θ \quad = \quad a^2\cosh^2θ\)

5)\(16\cosh^2θ−16\)

Tumia mbinu ya kukamilisha mraba kuelezea kila trinomial katika mazoezi 6 - 8 kama mraba wa binomial.

6)\(4x^2−4x+1\)

Jibu: \( 4(x−\frac{1}{2})^2\)

7)\(2x^2−8x+3\)

8)\(−x^2−2x+4\)

Jibu: \( −(x+1)^2+5\)

Katika mazoezi 9 - 28, kuunganisha kutumia njia ya kubadilisha trigonometric. Eleza jibu la mwisho kwa suala la kutofautiana awali.

9)\(\displaystyle ∫\frac{dx}{\sqrt{4−x^2}}\)

10)\(\displaystyle ∫\frac{dx}{\sqrt{x^2−a^2}}\)

Jibu: \(\displaystyle ∫\frac{dx}{\sqrt{x^2−a^2}} \quad = \quad \ln∣x+\sqrt{−a^2+x^2}∣+C\)

11)\(\displaystyle ∫\sqrt{4−x^2}\,dx\)

12)\(\displaystyle ∫\frac{dx}{\sqrt{1+9x^2}}\)

Jibu: \(\displaystyle ∫\frac{dx}{\sqrt{1+9x^2}} \quad = \quad \tfrac{1}{3}\ln∣\sqrt{9x^2+1}+3x∣+C\)

13)\(\displaystyle ∫\frac{x^2\,dx}{\sqrt{1−x^2}}\)

14)\(\displaystyle ∫\frac{dx}{x^2\sqrt{1−x^2}}\)

Jibu: \(\displaystyle ∫\frac{dx}{x^2\sqrt{1−x^2}} \quad = \quad −\frac{\sqrt{1−x^2}}{x}+C\)

15)\(\displaystyle ∫\frac{dx}{(1+x^2)^2}\)

16)\(\displaystyle ∫\sqrt{x^2+9}\,dx\)

Jibu: \(\displaystyle ∫\sqrt{x^2+9}\,dx \quad = \quad 9\left[\frac{x\sqrt{x^2+9}}{18}+\tfrac{1}{2}\ln\left|\frac{\sqrt{x^2+9}}{3}+\frac{x}{3}\right|\right]+C\)

17)\(\displaystyle ∫\frac{\sqrt{x^2−25}}{x}\,dx\)

18)\(\displaystyle ∫\frac{θ^3}{\sqrt{9−θ^2}}\,dθ\)

Jibu: \(\displaystyle ∫\frac{θ^3dθ}{\sqrt{9−θ^2}}\,dθ \quad = \quad −\tfrac{1}{3}\sqrt{9−θ^2}(18+θ^2)+C\)

19)\(\displaystyle ∫\frac{dx}{\sqrt{x^6−x^2}}\)

20)\(\displaystyle ∫\sqrt{x^6−x^8}\,dx\)

Jibu: \(\displaystyle ∫\sqrt{x^6−x^8}\,dx \quad = \quad \frac{(−1+x^2)(2+3x^2)\sqrt{x^6−x^8}}{15x^3}+C\)

21)\(\displaystyle ∫\frac{dx}{(1+x^2)^{3/2}}\)

22)\(\displaystyle ∫\frac{dx}{(x^2−9)^{3/2}}\)

Jibu: \(\displaystyle ∫\frac{dx}{(x^2−9)^{3/2}} \quad = \quad −\frac{x}{9\sqrt{x^2-9}}+C\)

23)\(\displaystyle ∫\frac{\sqrt{1+x^2}}{x}\,dx\)

24)\(\displaystyle ∫\frac{x^2}{\sqrt{x^2−1}}\,dx\)

Jibu: \(\displaystyle ∫\frac{x^2}{\sqrt{x^2−1}}\,dx \quad = \quad \tfrac{1}{2}(\ln∣x+\sqrt{x^2−1}∣+x\sqrt{x^2−1})+C\)

25)\(\displaystyle ∫\frac{x^2}{x^2+4}\,dx\)

26)\(\displaystyle ∫\frac{dx}{x^2\sqrt{x^2+1}}\)

Jibu: \(\displaystyle ∫\frac{dx}{x^2\sqrt{x^2+1}} \quad = \quad −\frac{\sqrt{1+x^2}}{x}+C\)

27)\(\displaystyle ∫\frac{x^2}{\sqrt{1+x^2}}\,dx\)

28)\(\displaystyle ∫^1_{−1}(1−x^2)^{3/2}\,dx\)

Jibu: \(\displaystyle ∫^1_{−1}(1−x^2)^{3/2}\,dx \quad = \quad \tfrac{1}{8}\left(x(5−2x^2)\sqrt{1−x^2}+3\arcsin x\right)+C\)

Katika mazoezi 29 - 34, tumia mbadala\(x=\sinh θ, \, \cosh θ,\) au Eleza\(\tanh θ.\) majibu ya mwisho kwa suala la kutofautiana\(x\).

29)\(\displaystyle ∫\frac{dx}{\sqrt{x^2−1}}\)

30)\(\displaystyle ∫\frac{dx}{x\sqrt{1−x^2}}\)

Jibu: \(\displaystyle ∫\frac{dx}{x\sqrt{1−x^2}} \quad = \quad \ln x−\ln∣1+\sqrt{1−x^2}∣+C\)

31)\(\displaystyle ∫\sqrt{x^2−1}\,dx\)

32)\(\displaystyle ∫\frac{\sqrt{x^2−1}}{x^2}\,dx\)

Jibu: \(\displaystyle ∫\frac{\sqrt{x^2−1}}{x^2}\,dx \quad = \quad −\frac{\sqrt{−1+x^2}}{x}+\ln\left|x+\sqrt{−1+x^2}\right|+C\)

33)\(\displaystyle ∫\frac{dx}{1−x^2}\)

34)\(\displaystyle ∫\frac{\sqrt{1+x^2}}{x^2}\,dx\)

Jibu: \(\displaystyle ∫\frac{\sqrt{1+x^2}}{x^2}\,dx \quad = \quad −\frac{\sqrt{1+x^2}}{x}+\text{arcsinh}\, x+C\)

Tumia mbinu ya kukamilisha mraba kutathmini integrals katika mazoezi 35 - 39.

35)\(\displaystyle ∫\frac{1}{x^2−6x}\,dx\)

36)\(\displaystyle ∫\frac{1}{x^2+2x+1}\,dx\)

Jibu: \(\displaystyle ∫\frac{1}{x^2+2x+1}\,dx \quad = \quad −\frac{1}{1+x}+C\)

37)\(\displaystyle ∫\frac{1}{\sqrt{−x^2+2x+8}}\,dx\)

38)\(\displaystyle ∫\frac{1}{\sqrt{−x^2+10x}}\,dx\)

Jibu: \(\displaystyle ∫\frac{1}{\sqrt{−x^2+10x}}\,dx \quad = \quad \arcsin\left( \frac{x-5}{5}\right)+C\)

39)\(\displaystyle ∫\frac{1}{\sqrt{x^2+4x−12}}\,dx\)

40) Tathmini muhimu bila kutumia calculus:\(\displaystyle ∫^3_{−3}\sqrt{9−x^2}\,dx.\)

Jibu: \(\displaystyle ∫^3_{−3}\sqrt{9−x^2}\,dx \quad = \quad \frac{9π}{2}\); eneo la semicircle na radius 3

41) Pata eneo lililofungwa na ellipse\(\dfrac{x^2}{4}+\dfrac{y^2}{9}=1.\)

42) Tathmini muhimu\(\displaystyle ∫\frac{dx}{\sqrt{1−x^2}}\) kwa kutumia mbadala mbili tofauti. Kwanza, hebu\(x=\cos θ\) na tathmini kutumia badala ya trigonometric. Pili, basi\(x=\sin θ\) na utumie badala ya trigonometric. Je, majibu ni sawa?

Jibu: \(\displaystyle ∫\frac{dx}{\sqrt{1−x^2}} \quad = \quad \arcsin(x)+C\)ni jibu la kawaida.

43) Tathmini muhimu\(\displaystyle ∫\frac{dx}{x\sqrt{x^2−1}}\) kwa kutumia badala\(x=\sec θ\). Kisha, tathmini muhimu sawa kwa kutumia ubadilishaji\(x=\csc θ.\) Onyesha kwamba matokeo ni sawa.

44) Tathmini muhimu\(\displaystyle ∫\frac{x}{x^2+1}\,dx\) kwa kutumia fomu\(\displaystyle ∫\frac{1}{u}\,du\). Kisha, tathmini muhimu sawa kutumia\(x=\tan θ.\) Je, matokeo ni sawa?

Jibu: \(\displaystyle ∫\frac{x}{x^2+1}\,dx \quad = \quad \frac{1}{2}\ln(1+x^2)+C\)ni matokeo kwa kutumia njia yoyote.

45) Eleza njia ya ushirikiano ungependa kutumia kutathmini muhimu\(\displaystyle ∫x\sqrt{x^2+1}\,dx.\) Kwa nini ulichagua njia hii?

46) Eleza njia ya ushirikiano ungependa kutumia kutathmini muhimu\(\displaystyle ∫x^2\sqrt{x^2−1}\,dx.\) Kwa nini ulichagua njia hii?

Jibu: Tumia badala ya trigonometric. Hebu\(x=\sec(θ).\)

47) Tathmini\(\displaystyle ∫^1_{−1}\frac{x}{x^2+1}\,dx\)

48) Pata urefu wa arc ya curve juu ya muda maalum:\(y=\ln x,\quad [1,5].\) Pindua jibu kwa maeneo matatu ya decimal.

Jibu: \( s = 4.367\)vitengo

49) Kupata eneo la uso wa imara yanayotokana na yanazunguka eneo imepakana na grafu ya\(y=x^2,\, y=0,\, x=0\), na\(x=\sqrt{2}\) kuhusu\(x\) -axis. (Pindua jibu kwa maeneo matatu ya decimal).

50) Kanda imepakana na grafu ya\(f(x)=\dfrac{1}{1+x^2}\) na\(x\) -axis kati\(x=0\) na\(x=1\) inahusu\(x\) -axis. Pata kiasi cha imara inayozalishwa.

Jibu: \( V = \left(\frac{π^2}{8}+\frac{π}{4}\right) \, \text{units}^3\)

Katika mazoezi 51 - 52, tatua tatizo la thamani ya awali kwa\(y\) kama kazi ya\(x\).

51)\((x^2+36)\dfrac{dy}{dx}=1, \quad y(6)=0\)

52)\((64−x^2)\dfrac{dy}{dx}=1, \quad y(0)=3\)

Jibu: \( y=\tfrac{1}{16}\ln\left|\dfrac{x+8}{x−8}\right|+3\)

53) Kupata eneo imepakana na\(y=\dfrac{2}{\sqrt{64−4x^2}},\, x=0,\, y=0\), na\(x=2\).

54) Tank ya kuhifadhi mafuta inaweza kuelezewa kama kiasi kilichozalishwa na kuzunguka eneo lililofungwa na\(y=\dfrac{16}{\sqrt{64+x^2}},\, x=0,\, y=0,\, x=2\) kuhusu\(x\) -axis. Pata kiasi cha tank (katika mita za ujazo).

Jibu: \(V = 24.6\)m ³

55) Wakati wa kila mzunguko, kasi\(v\) (kwa miguu kwa pili) ya kifaa cha kulehemu cha roboti hutolewa na\(v=2t−\dfrac{14}{4+t^2}\), wapi wakati\(t\) kwa sekunde. Kupata kujieleza kwa ajili ya makazi yao\(s\) (katika miguu) kama kazi ya\(t\) kama\(s=0\) wakati\(t=0\).

56) Kupata urefu wa Curve\(y=\sqrt{16−x^2}\) kati\(x=0\) na\(x=2\).

Jibu: \( s = \frac{2π}{3}\)vitengo