### Question

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

### Question

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]

Ans:

### Question

(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]

Ans:

(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 $$\frac{11}{96}$$

### Question

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. Answer/Explanation (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$$e-2+\frac{1}{e}$$

### Question

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 $$a(2x x^{2})e^{\frac{1}{2}x}+b\int (2-2x)e^{\frac{1}{2}x}dx$$

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

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

### Question

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