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STatiana [176]
3 years ago
5

(a) Use the reduction formula to show that integral from 0 to pi/2 of sin(x)^ndx is (n-1)/n * integral from 0 to pi/2 of sin(x)^

(n-2)dx where n>=2 is an integer.
(b)Use part (a) to evaluate integral from 0 to pi/2 of sin(x)^3dx and integral from 0 to pi/2 of sin(x)^5dx.

(c) Use part (a) to show that, for odd powers of sine, integral from 0 to pi/2 sin(x)^(2n 1)dx is (2*4*6*...*2n)/[3*5*7...*(2n 1)].
Mathematics
1 answer:
Sedbober [7]3 years ago
3 0
Hello,

a)
I= \int\limits^{ \frac{\pi}{2} }_0 {sin^n(x)} \, dx = \int\limits^{ \frac{\pi}{2} }_0 {sin(x)*sin^{n-1}(x)} \, dx \\

= [-cos(x)*sin^{n-1}(x)]_0^ \frac{\pi}{2}+(n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos(x)*sin^{n-2}(x)*cos(x)} \, dx \\

=0 + (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos^2(x)*sin^{n-2}(x)} \, dx \\

= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {(1-sin^2(x))*sin^{n-2}(x)} \, dx \\
= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx - (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^n(x) \, dx\\


I(1+n-1)= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\
I= \dfrac{n-1}{n} *\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\


b)
\int\limits^{ \frac{\pi}{2} }_0 {sin^{3}(x)} \, dx \\
= \frac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx \\
= \dfrac{2}{3}\ [-cos(x)]_0^{\frac{\pi}{2}}=\dfrac{2}{3} \\






\int\limits^{ \frac{\pi}{2} }_0 {sin^{5}(x)} \, dx \\
= \dfrac{4}{5}*\dfrac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx = \dfrac{8}{15}\\







c)

I_n=  \dfrac{n-1}{n} * I_{n-2} \\

I_{2n+1}=  \dfrac{2n+1-1}{2n+1} * I_{2n+1-2} \\
= \dfrac{2n}{2n+1} * I_{2n-1} \\
= \dfrac{(2n)*(2n-2)}{(2n+1)(2n-1)} * I_{2n-3} \\
= \dfrac{(2n)*(2n-2)*...*2}{(2n+1)(2n-1)*...*3} * I_{1} \\\\

I_1=1\\






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Answer:

The required recursive formula is

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Step-by-step explanation:

Mohamed decided to track the number of leaves on the tree in his backyard each year.

The first year there were 500 leaves

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Each year thereafter the number of leaves was 40% more than the year before so that means

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For the third year the number of leaves increase 40% than the year before so that means

Year \: 3 = 500\times 1.4(1+0.40) = 500 \times 1.4^{2}\\

Similarly for fourth year,

Year \: 4 = 500\times 1.4^{2}(1+0.40) = 500\times 1.4^{3}\\

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Let f(n) be the number of leaves on the tree in Mohameds back yard in the nth year since he started tracking it then general recursive formula is

f(n)= 500\times(1.4)^{n-1}\\

This is the required recursive formula to find the number of leaves for the nth year.

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Lets find out the number of leaves in the 10th year,

f(10)= 500\times(1.4)^{10-1}\\\\f(10)= 500\times(1.4)^{9}\\\\f(10)= 500\times20.66\\\\f(10)= 10330

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Complete question is;

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Now,the coordinates of the vertex will be (h, k)

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