二重积分求解,各位高手帮忙,最好详细点解答,谢谢
看不全,
1.A
2.C
3.C
4.B
5.D
6.A
解:
1.dare有两种词性:1.实义动词 2.情态动词
dare后面直接加的是not,可知dare在此作情态动词
若是实义动词就应该是The little boy does not dare to go to church.
所以就跟就反义疑问句原则:前否后肯,用dare he?
其实当你判断出它是情态动词时,你就可以把它替换成比较熟悉的can或者must等之类的词来分析了
2.此句为 You think 引导的宾语从句。选C
相关语法:带有定语从句或宾语从句的主从复合句,若其陈述部分为I(We) think(believe, suppose, consider, expect) + that从句及其否定形
式时,问句部分的动词及主语与that从句内的动词和主语保持一致。
如:① I think that he has done his best, hasn’t he? ② We think that English is very useful, isn’t it? (不用don’t we?)
③ I don’t think that you can do it, can you? (不用do I?) ④ We don’t believe that the news is true, is it? (不用do we?)
以上题不符合,所以以常规的来答
3.同样是宾语从句,问句部分的动词及主语与主句保持一致。 Nobody引导时,反义为肯定。所以用 will they,选C
相关语法:1.带有定语从句或宾语从句的主从复合句,问句部分的动词和主语与陈述部分的主句动词和主语保持一致
2.反意疑问句的陈述部分带有little, few, never, hardly, seldom , rarely , nothing , nobody 等否定意义的词时,问句部分用肯定式。
如: ①She never tells a lie, does she?(不用doesn’t she?) ②He was seldom late, was he?(不用wasn’t he?)
4. 这里的 must have done 表示推测,且没有时间状语。所以选B
相关语法:当用must(may, might) + have + V-ed表示推测时,若句中没有带明显的过去时间的状语,问句部分动词用现在完成时形式
如:①Everyone must have known the death of the waitress, haven’t they? (不用mustn’t they/ didn’t they ?)
②You must have worked there, haven’t you?(不用mustn’t you?/ didn’t you?)
5. 这里的must be意为一定是,反义用be(not)+主语,为aren't you,选D
相关语法:当must be表“一定、必定”时,疑问部分助动词就用be. 如:He must be very tired after running, isn’t he ?
6.此句为Let's 引导的祈使句,问句部分均用shall we,所以选A
相关语法:
陈述部分为祈使句时,除以Let’s……引导的问句部分shall we? 如: Let’s go home together, shall we? 外,其余祈使句(无论肯、否定),问句部分一般用will you?形式表示请求
如:①Let us stop to rest, will you? ②You feed the bird today, will you?
③Please open the window, will you? ④ Don’t make any noise, will you?
谢谢,望采纳。
积分区域会画吧
∫∫ F(x,y) = ∫[0,2] dx ∫ [x²,2x] x/(y+1) dy
= ∫[0,2] dx * xln(y+1) | [x²,2x]
= ∫[0,2] [ xln(2x+1) - xln(x²+1) ] dx
∫[0,2] xln(2x+1) dx
= 1/2 ∫[0,2] ln(2x+1) dx²
= 1/2 x²ln(2x+1) |[0,2] - 1/2 ∫ x² dln(2x+1)
= 2ln5 - ∫[0,2] x² / (2x+1) dx
= 2ln5 - ∫[0,2] [ (2x-1)/4 + 1/ (4(2x+1)) ] dx
= 2ln5 - [ (x²-x)/4 + 1/8 ln(2x+1) ] [0,2]
= 2ln5 - (1/2 + 1/8 ln5)
= 15/8 ln5 - 1/2
∫[0,2] xln(x²+1) dx
= 1/2 ∫[0,2] ln(x²+1) dx²
=1/2 x²ln(x²+1) |[0,2] - 1/2 ∫[0,2] x² dln(x²+1)
= 2ln5 - ∫[0,2] x³ / (x²+1) dx
= 2ln5 - ∫[0,2] [ x - x/(x²+1) ] dx
= 2ln5 - [ x²/2 - 1/2 ln(x²+1) ] [0,2]
= 2ln5 - (2 - 1/2 ln5)
= 5/2 ln5 - 2
∴原式 = 15/8 ln5 - 1/2 - (5/2 ln5 - 2) = (12 - 5ln5) /8
二重积分求解,各位高手帮忙,最好详细点解答,谢谢视频
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