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6. (i) Find \[ \int xe^{x} \: dx \] (ii) Find \[ \int \frac{8}{(2x-1)^{3}} \: dx, \, x > \frac{1}{2} \] (iii) Given that \( y = \frac{\pi}{6} \) at \( x = 0 \), solve the differential equation \[ \frac{dy}{dx} = e^{x} \: \csc(2y) \: \csc(y) \] - Edexcel - A-Level Maths Pure - Question 8 - 2014 - Paper 7
Question 8
![6.-(i)-Find-\[-\int-xe^{x}-\:-dx-\]--(ii)-Find-\[-\int-\frac{8}{(2x-1)^{3}}-\:-dx,-\,-x->-\frac{1}{2}-\]--(iii)-Given-that-\(-y-=-\frac{\pi}{6}-\)-at-\(-x-=-0-\),-solve-the-differential-equation-\[-\frac{dy}{dx}-=-e^{x}-\:-\csc(2y)-\:-\csc(y)-\]-Edexcel-A-Level Maths Pure-Question 8-2014-Paper 7.png](https://cdn.simplestudy.cloud/assets/backend/uploads/question/MathsPure_2014_7_1_QP_6_2014.jpg)
6. (i) Find \[ \int xe^{x} \: dx \] (ii) Find \[ \int \frac{8}{(2x-1)^{3}} \: dx, \, x > \frac{1}{2} \] (iii) Given that \( y = \frac{\pi}{6} \) at \( x = 0 \), so... show full transcript
Worked Solution & Example Answer:6. (i) Find \[ \int xe^{x} \: dx \] (ii) Find \[ \int \frac{8}{(2x-1)^{3}} \: dx, \, x > \frac{1}{2} \] (iii) Given that \( y = \frac{\pi}{6} \) at \( x = 0 \), solve the differential equation \[ \frac{dy}{dx} = e^{x} \: \csc(2y) \: \csc(y) \] - Edexcel - A-Level Maths Pure - Question 8 - 2014 - Paper 7
Step 1
Find \( \int xe^{x} \: dx \)
Answer
To find ( \int xe^{x} : dx ), we will use integration by parts. Let
- ( u = x ) and ( dv = e^{x} dx ), then ( du = dx ) and ( v = e^{x} ). By the integration by parts formula:
[ \int u , dv = uv - \int v , du ]
Substituting gives: [ \int xe^{x} : dx = xe^{x} - \int e^{x} : dx = xe^{x} - e^{x} + C = e^{x}(x - 1) + C ]
Step 2
Find \( \int \frac{8}{(2x-1)^{3}} \: dx, \, x > \frac{1}{2} \)
Answer
To solve this integral, we use the substitution method. Let
- ( u = 2x - 1 ), then ( du = 2dx ) or ( dx = \frac{1}{2} du ).
When ( x = \frac{1}{2} ), ( u = 0 ). The integral becomes: [ \int \frac{8}{u^{3}} \cdot \frac{1}{2} du = 4 \int u^{-3} du ]
Now integrating gives: [ 4 \cdot \left(-\frac{1}{2u^{2}}\right) + C = -\frac{2}{(2x-1)^{2}} + C ]
Step 3
Given that \( y = \frac{\pi}{6} \) at \( x = 0 \), solve the differential equation \( \frac{dy}{dx} = e^{x} \: \csc(2y) \: \csc(y) \)
Answer
We start with the differential equation: [ \frac{dy}{dx} = e^{x} : \csc(2y) : \csc(y) ]
Separate the variables: [ \int \csc(2y) , \csc(y) , dy = \int e^{x} , dx ]
Integrating both sides, let: [ \int \csc(2y) , \csc(y) , dy \text{ (this can be derived with appropriate substitutions) } ]\
The right side integrates to ( e^{x} + C ).
Now applying the initial condition ( y(0) = \frac{\pi}{6} ) can help find the constant. After finding ( C ), the solution will be complete.