Integration of Exponential Functions (1 Viewer)

V_L · Apr 3, 2017

Hi,

Could someone please explain the following questions:

9. Find the volume of the solid formed when the curve y = e^−x + 1 is rotated about the x -axis from x = 1 to x = 2, correct to 1 decimal place.

11. (b) find integral of x(2 + x)e^x dx.

12. The curve y = √(e^x + 1) is rotated about the x- axis from x = 0 to x = 1. Find the exact volume of the solid formed.

13. Find the exact area enclosed between the curve y = e^2x and the lines y = 1 and x = 2.

Answers:

9. 4.8 units^3
11. (b) x^2e^x + C
10. 7.4
12. πe units^3

integral95 · Apr 3, 2017

Question 9 and 12 basically uses the volume formula

$\pi \int_{a}^{b} y^2 \ \ dx$

Q11 b) can't be integrated with 2U knowledge unless there's a previous part given.

Q13 Drawing the graph out you get a part of an area under the curve but an open rectangle at the bottom you must subtract.

The working is

$\int_{0}^{2} e^{2x} \ dx - 2(1)$

i.e the area under the curve between x = 0 and 2 minus the rectangle

davidgoes4wce · Apr 3, 2017

$Q9.$

$V_{x}=\pi \int_{1}^2 y^2 dx$

$=\pi \int_{1}^2 (e^{-x}+1)^2 \ dx$

$=\pi \int_{1}^2 e^{-2x}+2e^{-x}+1 \ dx$

$= \pi [ \frac{-e^{-2x}}{2}-2e^{-x}+x] _{x=1} ^{x=2}$

$=\pi [( \frac{-e^{-4}}{2}-2e^{-2}+2)-( \frac{-e^{-2}}{2}-2e^{-1}+1]$

$= \pi [( \frac{-e^{-4}}{2}-2e^{-2}+2)+\frac{1}{2e^2}+\frac{2}{e}-1]$

$= \pi [\frac{-1}{2e^4}-\frac{3}{2e^2}+1+\frac{2}{e}]$

$\approx \ 4.8 units^{2}$

BenHowe · Apr 3, 2017

BenHowe · Apr 3, 2017

BenHowe · Apr 3, 2017

davidgoes4wce said:
$Q9.$

$V_{x}=\pi \int_{1}^2 y^2 dx$

$=\pi \int_{1}^2 (e^{-x}+1)^2 \ dx$

$=\pi \int_{1}^2 e^{-2x}+2e^{-x}+1 \ dx$

$= \pi [ \frac{-e^{-2x}}{2}-2e^{-x}+x] _{x=1} ^{x=2}$

$=\pi [( \frac{-e^{-4}}{2}-2e^{-2}+2)-( \frac{-e^{-2}}{2}-2e^{-1}+1]$

$= \pi [( \frac{-e^{-4}}{2}-2e^{-2}+2)+\frac{1}{2e^2}+\frac{2}{e}-1]$

$= \pi [\frac{-1}{2e^4}-\frac{3}{2e^2}+1+\frac{2}{e}]$

$\approx \ 4.8 units^{2}$

Dammit im too slow @ LaTeX RIP

1729 · Apr 3, 2017

V_L said:
Hi,

Could someone please explain the following questions:

9. Find the volume of the solid formed when the curve y = e^−x + 1 is rotated about the x -axis from x = 1 to x = 2, correct to 1 decimal place.

11. (b) find integral of x(2 + x)e^x dx.

12. The curve y = √(e^x + 1) is rotated about the x- axis from x = 0 to x = 1. Find the exact volume of the solid formed.

13. Find the exact area enclosed between the curve y = e^2x and the lines y = 1 and x = 2.

Answers:

9. 4.8 units^3
11. (b) x^2e^x + C
10. 7.4
12. πe units^3

$$\noindent \textbf{Q9.} $ V = \pi \int_1^2(e^{-x}+1)^2\mathrm{d}x = \pi \int_1^2(e^{-2x} + 2e^{-x} + 1)\mathrm{d}x = \pi \left[-\frac{1}{2}e^{-2x} -2e^{-x} + x \right]^2_1 = \pi \left(-\frac{1}{2}e^{-4} - 2e^{-2} + 2 + \frac{1}{2}e^{-2} + 2e^{-1} - 1\right) = 4.8$ units$^3$ (correct to 1 dp.)$

$$\noindent \textbf{Q11.}$ \int x(2+x)e^x\mathrm{d}x = \int (2xe^x + x^2e^x)\mathrm{d}x$. If we use logic, this is obviously a product rule application whereby $u = x^2$ and $v = e^x$, note $\frac{\mathrm{d}u}{\mathrm{d}x} = 2x.$ Thus, $\int x(2+x)e^x \mathrm{d}x = x^2e^x + C$, for some constant $C$.$

$\noindent \textbf{Q12.} $ V = \pi \int_0^1(\sqrt{e^x+1})^2\mathrm{d}x = \pi \int_0^1(e^x+1)\mathrm{d}x = \pi \left[e^x + x\right]^1_0 = \pi(e + 1 - e^0) = \pi e $ units$^3$

$\noindent \textbf{Q13.} Upon graphical inspection, we see that $A = \int_0^2e^{2x}\mathrm{d}x - 2 \times 1$. Note that the integral we used represents the required area along with an unnecessary rectangle that we must subtract. Thus, $A = \left[\frac{1}{2}e^{2x}\right]^2_0 -2 = \frac{1}{2}e^4 - \frac{1}{2} - 2 = \frac{1}{2}e^4 - \frac{5}{2} = \frac{1}{2}(e^4 - 5)$ units$^2$

BenHowe · Apr 3, 2017

1729 said:
$$\noindent \textbf{Q11.}$ \int x(2+x)e^x\mathrm{d}x = \int (2xe^x + x^2e^x)\mathrm{d}x$. If we use logic, this is obviously a product rule application whereby $u = x^2$ and $v = e^x$, note $\frac{\mathrm{d}u}{\mathrm{d}x} = 2x.$ Thus, $\int x(2+x)e^x \mathrm{d}x = x^2e^x + C$, for some constant $C$.$

... i always forget to check for reverse product and reverse quotient

V_L · Apr 3, 2017

Thankyou !!

davidgoes4wce · Apr 4, 2017

No worries mate!

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Integration of Exponential Functions (1 Viewer)

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