5. (i) Find, by algebraic integration, the exact value of
$$\int _ { 2 } ^ { 4 } \frac { 8 } { ( 2 x - 3 ) ^ { 3 } } d x$$
(ii) Find, in simplest form,
$$\int x \left( x ^ { 2 } + 3 \right) ^ { 7 } d x$$
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Question 5:
Part (i):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(\int \frac{8}{(2x-3)^3}\,dx = \frac{-2}{(2x-3)^2}(+c)\) M1 A1
M1: achieves \(\frac{A}{(2x-3)^2}\); A1: achieves \(\frac{-2}{(2x-3)^2}\) (indices must be processed; no requirement for \(+c\))
\(\int_2^4 \frac{8}{(2x-3)^3}\,dx = \left[\frac{-2}{(2x-3)^2}\right]_2^4 = -\frac{2}{25}+2 = \frac{48}{25}\) dM1 A1
dM1: substitutes 2 and 4, subtracts (either way); A1: \(\frac{48}{25}\) or 1.92 isw
Part (ii):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(\int x(x^2+3)^7\,dx = \frac{1}{16}(x^2+3)^8 + c\) M1 A1
M1: achieves \(k(x^2+3)^8\); A1: \(\frac{1}{16}(x^2+3)^8 + c\), must be in terms of \(x\), \(+c\) must be present
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## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{8}{(2x-3)^3}\,dx = \frac{-2}{(2x-3)^2}(+c)$ | M1 A1 | M1: achieves $\frac{A}{(2x-3)^2}$; A1: achieves $\frac{-2}{(2x-3)^2}$ (indices must be processed; no requirement for $+c$) |
| $\int_2^4 \frac{8}{(2x-3)^3}\,dx = \left[\frac{-2}{(2x-3)^2}\right]_2^4 = -\frac{2}{25}+2 = \frac{48}{25}$ | dM1 A1 | dM1: substitutes 2 and 4, subtracts (either way); A1: $\frac{48}{25}$ or 1.92 isw |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int x(x^2+3)^7\,dx = \frac{1}{16}(x^2+3)^8 + c$ | M1 A1 | M1: achieves $k(x^2+3)^8$; A1: $\frac{1}{16}(x^2+3)^8 + c$, must be in terms of $x$, $+c$ must be present |
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5. (i) Find, by algebraic integration, the exact value of
$$\int _ { 2 } ^ { 4 } \frac { 8 } { ( 2 x - 3 ) ^ { 3 } } d x$$
(ii) Find, in simplest form,
$$\int x \left( x ^ { 2 } + 3 \right) ^ { 7 } d x$$
\hfill \mbox{\textit{Edexcel P3 2021 Q5 [6]}}