Can someone please check my solutions to 7a and 22b (1 Viewer)

kpad5991 · Jan 12, 2018

https://imgur.com/gallery/z7tsn

Can someone please send me solutions cause I'm not getting the same answer as the book to 7a and is that the correct integral for 22b can you please integrate that if you can please. I'm stuck

InteGrand · Jan 12, 2018

kpad5991 said:
https://imgur.com/gallery/z7tsn

Can someone please send me solutions cause I'm not getting the same answer as the book to 7a and is that the correct integral for 22b can you please integrate that if you can please. I'm stuck

For 7a), that's not the right integral for the area of the given region. Your integral finds the area under the exponential curve, but you want the area between a horizontal line and the exponential curve.

A hint to get the right answer for 7a) without doing any more integrating: the desired area is the area of a certain rectangle minus the area under the exponential curve.

InteGrand · Jan 12, 2018

kpad5991 said:
https://imgur.com/gallery/z7tsn

Can someone please send me solutions cause I'm not getting the same answer as the book to 7a and is that the correct integral for 22b can you please integrate that if you can please. I'm stuck

For 22b), you have made an error in the squaring. Double check the exponential term (and your index laws if need be).

fan96 · Jan 12, 2018

7a) The integral $\int^2_0 e^{-x} dx$ gives the area UNDER the curve. You need to subtract that from 2 (area of rectangle) to get the correct answer.

$2 - \int^2_0 e^{-x} dx \\ \\= 2 - [-e^{-x}]^2_0 \\\\ = 2 - (-e^{-2} - (- 1)) \\\\ = 1 + e^{-2}$

22b)
$A = \pi \int^2_0 (\sqrt{2x} e^{-\frac{1}{2}x^2})^2 dx \\\\ = \pi \int^2_0 2x (e^{-\frac{1}{2}x^2})^2 \\\\ \text{Note:}\, (x^m)^n = x^{mn} \\\\ = \pi \int^2_0 2x e^{-x^2} \\\\= -\pi \int^2_0 -2x e^{-x^2} \\\\ = -\pi \left[e^{-x^2}\right]^2_0\,\,\, \text{(From 22a)} \\\\ = -\pi\left( e^{-4} - 1 \right) \\\\ = 3.084\, \text{cm}^3\,\, \text{(4 s.f.)}$

captainhelium · Jan 12, 2018

$For question 7a, I think the integral should be$

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

$The line y=1 is on top of y= e^{-x}.$ $When you have two functions, the area enclosed is basically$

$\int_{a}^{b} Top - Bottom$

$So solving the integral, we get$

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

$x + e^{-x}\Biggr|_{0}^{2}$

$$ ∴ 2+ e^{-2} -1$

$∴ Answer is 1 + e^{-2}$

$For Q22b, I think the integral should instead be$

$\pi $ \int_{0} ^ {2} 2xe^{-x^2} $$

$Just use integration via substitution. Let u= -x^2.$ $Or just use your answer from part a)$

I hope everything was correct lol. I'm new to LaTeX input so I hope I didn't make a mistake haha

Can someone please check my solutions to 7a and 22b (1 Viewer)

kpad5991

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InteGrand

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InteGrand

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fan96

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captainhelium

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