Show that \(\int_{1}^{4}x^-\frac{3}{2}\)In \(xdx=2-In4\).

Integrate by parts and reach \(ax^{\tfrac{1}{2}}Inx+b\int x^{\frac{-1}{2}}.\frac{1}{2}dx\) Obtain \(-2x^{\frac{-1}{2}}In x+2\int x^{\frac{-1}{2}}.\frac{1}{x}dx\), or equivalent Complete the integration, obtaining \(-2x^{\frac{1}{2}}Inx-4\)


\(x^{\frac{1}{2}}\),or equivalent Substitute limits correctly, having integrated twice  Obtain the given answer following full and correct working


The parametric equations of a curve are

x = e−t cost, y = e−t sin t.

    Show that \(\frac{dy}{dx}= tan \left ( t-\frac{1}{4}\pi \right )\) .[6]




 (a) Show that \(\int_{2}^{4}4xlnxdx=56ln2-12\). [5]

    (b) Use the substitution u = sin 4x to find the exact value of \(\int_{0}^{\frac{1}{24}\pi }cos^{3}4xdx.\) [5]



 (a) Carry out integration by parts and reach \(ax^{2}lnx+b\int \frac{1}{2}x^{2}dx\)
           Obtain \(2x^{2}lnx-\int \frac{1}{x}.2x^{2}dx\)
           Obtain \(2x^{2}lnx-x^{2}\)
           Use limits, having integrated twice 
           Confirm given result 56 ln 2 – 12

   (b) State or imply \(\frac{du}{dx}=4cos4x\)
           Carry out complete substitution except limits
           Obtain \(\int \left ( \frac{1}{4}- \frac{1}{4}u^{2}\right )du\) or equivalent
           Integrate to obtain form k1u + k2u3 with non-zero constants k1, k2
           Use appropriate limits to obtain \(\frac{11}{96}\)


The diagram shows the curve \(y =(Inx)^{2}\)

The x-coordinate of the point P is equal to e, and the normal to the curve at P meets the x-axis at Q.
(i) Find the x-coordinate of Q.

(ii) Show that\( \int In x dx  = x ln x − x + c, where c is a constant.

(iii) Using integration by parts, or otherwise, find the exact value of the area of the shaded region
between the curve, the x-axis and the normal PQ.


(1)  State or imply derivative is\( 2\frac{Inx}{x}\)

State or imply gradient of the normal at x =\( e is -\frac{1}{2}e,\)or equivalent 

 Carry out a complete method for finding the x-coordinate of Q

Obtain answer\( x=e+\frac{2}{e}\) exact equivalent 

(ii)  Justify the given statement by integration or by differentiation 

(iii)  Integrate by parts and reach  \(ax(Inx)^{2}+b\int x.\frac{Inx}{x}dx\) Complete the integration and obtain x(In x)2 – 2x In x + 2x, or equivalent

use limits x = 1 and x = e correctly, having integrated twice

Obtain exact value e-2 

Use x-coordinate of Q found in part (i) and obtain final answer\( e-2+\frac{1}{e}\) 


The diagram shows part of the curve\( y =(2x-x^{2})e^{\frac{1}{2}x}\) and its maximum point M.

(i) Find the exact x-coordinate of M. 
(ii) Find the exact value of the area of the shaded region bounded by the curve and the positive x-axis



(i) Use the correct product rule
Obtain correct derivative in any form, e.g.\( (2-2x)e^{\frac{1}{2}}+\frac{1}{2}(2x-x^{2})e^{\frac{1}{2}x}\)

Equate derivative to zero and solve for x

Obtain\( x \sqrt{5-1}\)

(ii) Integrate by parts and reach \(a(2x x^{2})e^{\frac{1}{2}x}+b\int (2-2x)e^{\frac{1}{2}x}dx\)

\(2e^{\frac{1}{2}x}(2x-x^{2})-2\int (2-2x)e^{\frac{1}{2}x}\)

Complete the integration correctly, obtaining\( (12x-2x^{2}-24)e^{\frac{1}{2}}\)

Use limits x = 0, x = 2 correctly having integrated by parts twice
Obtain answer 24 – 8e, or exact simplified equivalent


Find the exact value of \(\int_{0}^{\frac{1}{2}} xe^{-2x}dx\)

Integrate by parts and reach \(axe^{-2x}+b\int e^{-2x}\)

Obtain \(\frac{1}{2}xe^{-2x}+\frac{1}{2},\int e^{-2x}\)dx  or equivalent

Complete the integration correctly, obtaining\( -\frac{1}{2}xe^{-2x}\)


\(-\frac{1}{4}e^{-2x}\),or equivalent

Use limits x = 0 and \(x=\frac{1}{2} \)correctly, having integrated twice

Obtain answer\( \frac{1}{4}-\frac{1}{2}e^{-1}\)  , or exact equivalent