Home / CIE A level -Pure Mathematics 2 : Topic :  2.5 Integration- integration : Exam Style Questions Paper 2

Question

Use the trapezium rule with 3 intervals to estimate the value of \(\int_{0}^{3}|2^{x}-4|dx\)

Answer/Explanation

State or imply ordinates 3,2,0,4

Use correct formula,or equivalent,with h=1 and four ordinates

Obtain answer 5.5.

Question

     

      The diagram shows the curve y = 2 + e−2x
      The curve crosses the y-axis at the point A, and the point B
      on the curve has x-coordinate 1. The shaded region is bounded by the curve and the line segment AB.
      Find the exact area of the shaded region.                                                                                                               [5]

Answer/Explanation

Ans

3 State \(\frac{dx}{dt}-e^{t} + 2e^{-t}, \frac{dy}{dt}=6e^{2t}\)

    \(\frac{dy}{dx}=\frac{dy}{dt}|\frac{dx}{dt}\) either in terms of t or after substitution of t = 0 

    Obtain gradient of tangent is 2 A1
    Attempt equation of tangent with numerical gradient and coordinates 
    Obtain y = 2x + 6 or equivalent 

Question

It is given that \(\int_{a}^{3a}\frac{2}{2x-5}dx=In\frac{7}{2}\).
Find the value of the positive constant a.

Answer/Explanation

Ans:

Integrate to obtain k In(2x-5)
Apply limits to obtain In(6a-5)-In(2a-5)=In\(\frac{7}{2}\)
Apply subtraction law for logarithms
Obtain equation\(\frac{6a-5}{2a-5}=\frac{7}{2}\)
Solve equation for a
Obtain \(a=\frac{25}{2}\)

Question

(a) Express \(5\sqrt{3}cos x + 5sin x\) in the form \(R cos(x- \alpha )\), where R>0 and \(0<\alpha<\frac{1}{2}\pi\).
(b) As x varies, find the least possible value of
\(4+5\sqrt{3}cos x+5sinx\),
and determine the corresponding value of x where \(-\pi<x<\pi\).
(c) Find \(\int \frac{1}{(5\sqrt{3}cos3\theta+5sin3\theta)^2}d\theta\)

Answer/Explanation

Ans:

  1. State R=10
    Use appropriate trigonometry to find \(\alpha\)
    Obtain \(\alpha=\frac{1}{6}\pi\)
  2. State -6
    Attempt to find x from their \(cos(x-\alpha)=-1\)
    Obtain \(x-\frac{1}{6}\pi=-\pi\) and hence \(-\frac{5}{6}\pi\)
  3. State integrand of form \(k_1sec^2(3\theta-\frac{1}{6}\pi)\)
    Integrate to obtain form \(k_2tan(3\theta-\frac{1}{6}\pi)\)
    Obtain \(\frac{1}{300}tan(3\theta-\frac{1}{6}\pi)+c\)
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