ぐらぐらインテグラル⑤

ぐらぐらインテグラル⑤

11th - 12th Grade

8 Qs

quiz-placeholder

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ぐらぐらインテグラル⑤

ぐらぐらインテグラル⑤

Assessment

Quiz

Mathematics

11th - 12th Grade

Practice Problem

Hard

Created by

Kenji Takei

Used 10+ times

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8 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt


 xnx^n  の不定積分の公式を選べ

 nn  は 00  以上の整数)

 xnbx=1n+1xn+1+C\int_{ }^{ }x^nbx=\frac{1}{n+1}x^{n+1}+C  

 xndx=1n1xn1+C\int_{ }^{ }x^ndx=\frac{1}{n-1}x^{n-1}+C  

 xndx=1n+1xn+1+C\int_{ }^{ }x^ndx=\frac{1}{n+1}x^{n+1}+C  

 xndx=1n+1xn1+C\int_{ }^{ }x^ndx=\frac{1}{n+1}x^{n-1}+C  

この中にはない

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 (43x)2dx\int_{ }^{ }\left(4-3x\right)^2dx  を求めよ(積分定数C)

(ヒント: 1a (ax+b)n+1n+1+C\frac{1}{a}\ \frac{\left(ax+b\right)^{n+1}}{n+1}+C  )

 19(3x+4)3+C\frac{1}{9}\left(3x+4\right)^3+C  

 19(4x3)3+C-\frac{1}{9}\left(4x-3\right)^3+C  

 19(3x4)3+C\frac{1}{9}\left(3x-4\right)^3+C  

 13(3x2)3+C-\frac{1}{3}\left(3x-2\right)^3+C  

この中に答えはない

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 x(x1)dx\int_{ }^{ }x\left(x-1\right)dx  を求めよ

(積分定数C)

 13x312x2+C\frac{1}{3}x^3-\frac{1}{2}x^2+C  

 13x3+12x2+C\frac{1}{3}x^3+\frac{1}{2}x^2+C  

 x3+x2+Cx^3+x^2+C  

 x312x2+Cx^3-\frac{1}{2}x^2+C  

この中に答えはない

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 f(x)=2x2f'\left(x\right)=2x^2  , f(3)=0f\left(3\right)=0  を満たす関数 f(x)f\left(x\right)  を求めよ



 f(x)=23x318f\left(x\right)=\frac{2}{3}x^3-18  

 f(x)=23x3+18f\left(x\right)=\frac{2}{3}x^3+18  

 f(x)=x318f\left(x\right)=x^3-18  

 f(x)=x3+18f\left(x\right)=x^3+18  

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 (x1)(3x+1)dx\int_{ }^{ }\left(x-1\right)\left(3x+1\right)dx  を求めよ。

(積分定数をCとする)



 x3x2x+Cx^3-x^2-x+C  

 13x3x2x+C\frac{1}{3}x^3-x^2-x+C  

 x3x2+x+Cx^3-x^2+x+C  

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 (t+2)2dt\int_{ }^{ }\left(t+2\right)^2dt  を求めよ

(積分定数をCとする)

 13t3+2t2+4t+C\frac{1}{3}t^3+2t^2+4t+C  

 13t2+2t2+4t+C\frac{1}{3}t^2+2t^2+4t+C  

 3t3+2t2+4t+C3t^3+2t^2+4t+C  

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 (3x256)3dx\int_{ }^{ }\left(3x-256\right)^3dx  を求めよ

(積分定数はCとする)

 116(3x256)4\frac{1}{16}\left(3x-256\right)^4  

 112(3x256)4\frac{1}{12}\left(3x-256\right)^4  

 112(3x256)3+C\frac{1}{12}\left(3x-256\right)^3+C  

 112(3x256)4+C\frac{1}{12}\left(3x-256\right)^4+C  

 116(3x256)3+C\frac{1}{16}\left(3x-256\right)^3+C  

8.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 4x34x^3  の原始関数を選べ



 12x3+1412x^3+\frac{1}{4}  

 12x338612x^3-386  

 16x3+232116x^3+\frac{23}{21}  

 16x3+30016x^3+300  

この中にはない