9 Find the exact value of \(\int _ { 0 } ^ { 1 } \left( x ^ { 2 } + 1 \right) \mathrm { e } ^ { 2 x } \mathrm {~d} x\).
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Question 9:
Answer Marks
Guidance
Answer Marks
Guidance
Use \(u = x^2 + 1\), \(dv = e^{2x}\) or \(u = x^2\), \(dv = e^{2x}\) M1
\(1^{\text{st}}\) stage \(= f(x) +/- \int g(x)\, dx\)
\(1^{\text{st}}\) stage \(= \frac{1}{2}(x^2+1)e^{2x} - \int x\,e^{2x}\,dx\) or \(\frac{1}{2}x^2e^{2x} - \int xe^{2x}\,dx\) A1
For \(\int x\,e^{2x}\,dx\), use \(u = x\), \(dv = e^{2x}\) M1
ditto
\(= \frac{1}{2}x\,e^{2x} - \frac{1}{4}e^{2x}\) A1
Tolerate second sign error in \(-\int xe^{2x}\,dx\)
Complete final stage \(= \frac{1}{2}(x^2+1)e^{2x} - \frac{1}{4}(2x-1)e^{2x}\) A1
soi; may be separate terms
Correct (method) use of limits seen anywhere M1
Do not accept \((\ldots) - 0\)
Final answer \(= \frac{3}{4}e^2 - \frac{3}{4}\) A1
ISW; if A0, answer of \(4.79\ldots \rightarrow\) M1
[7]
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## Question 9:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $u = x^2 + 1$, $dv = e^{2x}$ or $u = x^2$, $dv = e^{2x}$ | M1 | $1^{\text{st}}$ stage $= f(x) +/- \int g(x)\, dx$ |
| $1^{\text{st}}$ stage $= \frac{1}{2}(x^2+1)e^{2x} - \int x\,e^{2x}\,dx$ or $\frac{1}{2}x^2e^{2x} - \int xe^{2x}\,dx$ | A1 | |
| For $\int x\,e^{2x}\,dx$, use $u = x$, $dv = e^{2x}$ | M1 | ditto |
| $= \frac{1}{2}x\,e^{2x} - \frac{1}{4}e^{2x}$ | A1 | Tolerate second sign error in $-\int xe^{2x}\,dx$ |
| Complete final stage $= \frac{1}{2}(x^2+1)e^{2x} - \frac{1}{4}(2x-1)e^{2x}$ | A1 | soi; may be separate terms |
| Correct (method) use of limits seen anywhere | M1 | Do not accept $(\ldots) - 0$ |
| Final answer $= \frac{3}{4}e^2 - \frac{3}{4}$ | A1 | ISW; if A0, answer of $4.79\ldots \rightarrow$ M1 |
| **[7]** | | |
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9 Find the exact value of $\int _ { 0 } ^ { 1 } \left( x ^ { 2 } + 1 \right) \mathrm { e } ^ { 2 x } \mathrm {~d} x$.
\hfill \mbox{\textit{OCR C4 2012 Q9 [7]}}