In a geometric progression (G.P), the first term is termed (a), and the common ratio is denoted by (r). Hence, the nth term is calculated using the formula;

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Given that;

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Now ,substituting the given values;

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7 0

1 year ago

Apply Gaussian quadrature with n = 4 to approximate integrate sin x^2dx from 1 to 5

maks197457 [2]

First, recall that Gaussian quadrature is based around integrating a function over the interval [-1,1], so transform the function argument accordingly to change the integral over [1,5] to an equivalent one over [-1,1].

$x=2t+3\iff t=\dfrac x2-\dfrac32\implies2\mathrm dt=\mathrm dx$
$x=1\implies t=\dfrac{2-6}4=-1$
$x=5\implies t=\dfrac{10-6}4=1$

So,

$\displaystyle\int_{x=1}^{x=5}\sin x^2\,\mathrm dx=\displaystyle2\int_{t=-1}^{t=1}\sin(2t+3)^2\,\mathrm dt$

Let $f(t)=2\sin(2t+3)^2$ . With $n=4$ , we're looking for coefficients $c_i$ and nodes $x_i$ , with $1\le i\le4$ , such that

$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx c_1f(x_1)+\cdots+c_4f(x_4)$

You can either try solving for each with the help of a calculator, or look up the values of the weights and nodes (they're extensively tabulated, and I'll include a link to one such reference).

Using the quadrature, we then have

$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.3749f(-0.8611)+0.6521f(-0.3400)+0.6521f(0.3400)+0.3749f-0.8611)$
$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.5790$

4 0

3 years ago

Nico completes 1/8 of his math problems in 2/5 of an hour.At tis rate how many of his math problems will he he finish in 1 hour

Whitepunk [10]

Answer:

it is 1/8÷2/5= 5/16

Step-by-step explanation:

did that help??????????????????

7 0

3 years ago