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AP Intermediate 1st Year Probability & Statistics — practice questions — page 2

Explore 34 free Probability & Statistics questions in AP Intermediate 1st Year Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Probability & Statistics questions and exercises with answers

Showing questions 31–34 of 34.

$\int \dfrac{x\,dx}{(x-1)(x-2)}$ equalsRead the answer and worked solution →$\int \dfrac{dx}{x(x^2+1)}$ equalsRead the answer and worked solution →The integration-by-parts formula $\int u\,\dfrac{dv}{dx}\,dx$ equalsRead the answer and worked solution →$\int x\cos x\,dx$ equalsRead the answer and worked solution →

Worked practice questions

$\int \dfrac{x\,dx}{(x-1)(x-2)}$ equals

  1. $\log\left|\dfrac{(x-1)^2}{x-2}\right|+C$
  2. $\log\left|\dfrac{(x-2)^2}{x-1}\right|+C$
  3. $\log\left|\left(\dfrac{x-1}{x-2}\right)^2\right|+C$
  4. $\log|(x-1)(x-2)|+C$
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Correct answer: B. $\log\left|\dfrac{(x-2)^2}{x-1}\right|+C$

1. Partial fractions: $\dfrac{x}{(x-1)(x-2)}=\dfrac{A}{x-1}+\dfrac{B}{x-2}$. 2. $x=1$: $1=A(-1)$ so $A=-1$. $x=2$: $2=B(1)$ so $B=2$. 3. Integrate: $-\log|x-1|+2\log|x-2|$. 4. Combine: $\log\left|\dfrac{(x-2)^2}{x-1}\right|+C$. Option A swaps the powers. _Source: syllabus Mathematics Ch 7 "Integrals", p.322_

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$\int \dfrac{dx}{x(x^2+1)}$ equals

  1. $\log|x|-\dfrac{1}{2}\log(x^2+1)+C$
  2. $\log|x|+\dfrac{1}{2}\log(x^2+1)+C$
  3. $-\log|x|+\dfrac{1}{2}\log(x^2+1)+C$
  4. $\dfrac{1}{2}\log|x|+\log(x^2+1)+C$
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Correct answer: A. $\log|x|-\dfrac{1}{2}\log(x^2+1)+C$

1. Write $\dfrac{1}{x(x^2+1)}=\dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}$. 2. Solving: $A=1,\ B=-1,\ C=0$, so integrand $=\dfrac{1}{x}-\dfrac{x}{x^2+1}$. 3. $\int \dfrac{1}{x}\,dx=\log|x|$; $\int \dfrac{x}{x^2+1}\,dx=\dfrac{1}{2}\log(x^2+1)$. 4. Result: $\log|x|-\dfrac{1}{2}\log(x^2+1)+C$. Option C flips the first sign. _Source: syllabus Mathematics Ch 7 "Integrals", p.322_

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The integration-by-parts formula $\int u\,\dfrac{dv}{dx}\,dx$ equals

  1. $uv+\displaystyle\int v\,\dfrac{du}{dx}\,dx$
  2. $uv-\displaystyle\int v\,\dfrac{du}{dx}\,dx$
  3. $\displaystyle\int v\,\dfrac{du}{dx}\,dx-uv$
  4. $uv-\displaystyle\int u\,\dfrac{du}{dx}\,dx$
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Correct answer: B. $uv-\displaystyle\int v\,\dfrac{du}{dx}\,dx$

1. From the product rule $\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$. 2. Integrating both sides: $uv=\int u\dfrac{dv}{dx}\,dx+\int v\dfrac{du}{dx}\,dx$. 3. Rearranging gives $\int u\dfrac{dv}{dx}\,dx=uv-\int v\dfrac{du}{dx}\,dx$. _Source: syllabus Mathematics Ch 7 "Integrals", p.323_

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