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Free_Kalibri [48]
2 years ago
13

If the graph of y=f(x) passes through the point (0,1), and dy/dx=xsin(x^2)/y, then f(x)= ?

Mathematics
1 answer:
Zigmanuir [339]2 years ago
4 0

The differential equation

d<em>y</em>/d<em>x</em> = <em>x</em> sin(<em>x</em> ²) / <em>y</em>

is separable as

<em>y</em> d<em>y</em> = <em>x</em> sin(<em>x</em> ²) d<em>x</em>

Integrate both sides:

∫ <em>y</em> d<em>y</em> = ∫ <em>x</em> sin(<em>x</em> ²) d<em>x</em>

∫ <em>y</em> d<em>y</em> = 1/2 ∫ 2<em>x</em> sin(<em>x</em> ²) d<em>x</em>

∫ <em>y</em> d<em>y</em> = 1/2 ∫ sin(<em>x</em> ²) d(<em>x</em> ²)

1/2 <em>y</em> ² = -1/2 cos(<em>x</em> ²) + <em>C</em>

Solve for <em>y</em> implicitly:

<em>y</em> ² = -cos(<em>x</em> ²) + <em>C</em>

Given that <em>y</em> = 1 when <em>x</em> = 0, we get

1² = -cos(0²) + <em>C</em>

1 = -cos(0) + <em>C</em>

1 = -1 + <em>C</em>

<em>C</em> = 2

Then the particular solution to the DE is

<em>y</em> ² = 2 - cos(<em>x</em> ²)

Solving explicitly for <em>y</em> would give two solutions,

<em>y</em> = ± √(2 - cos(<em>x</em> ²))

but only the one with the positive square root satisfies the initial condition:

√(2 - cos(0²)) = √1 = 1

-√(2 - cos(0²)) = -√1 = -1 ≠ 1

So the unique solution to this DE is

<em>y</em> = √(2 - cos(<em>x</em> ²))

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From the information given, it should be noted that DBC isn't an isosceles triangle.

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Likewise, BD and DC aren't perpendicular.

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express the limit as a definite integral on the given interval. lim n → [infinity] n ∑ i = 1 cos x i x i δ x , [ 2 π , 4 π ]
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The limit as a definite integral on the interval $\lim _{n \rightarrow \infty} \sum_{i=1}^n \frac{\cos x_i}{x_i} \Delta x$ on [2π , 4π] is $\int_{2\pi}^{4 \pi} \frac{\cos x}{x} d x$$.

<h3>What is meant by definite integral?</h3>

A definite integral uses infinitesimal slivers or stripes of the region to calculate the area beneath a function. Integrals can be used to represent a region's (signed) area, the cumulative value of a function changing over time, or the amount of a substance given its density.

Definite integral, a term used in mathematics. is the region in the xy plane defined by the graph of f, the x-axis, and the lines x = a and x = b, where the area above the x-axis adds to the total and the area below the x-axis subtracts from the total.

If an antiderivative F exists for the interval [a, b], the definite integral of the function is the difference of the values at points a and b. The definite integral of any function can also be expressed as the limit of a sum.

Let the equation be

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substitute the values in the above equation, we get

= $\lim _{n \rightarrow \infty} \sum_{i=1}^n \frac{\cos x_i}{x_i} \Delta x$ on [2π, 4π],

simplifying the above equation

$\int_{2\pi}^{4 \pi} \frac{\cos x}{x} d x$$

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Find the central angle of a sector of a circle if the area of the sector and the area of the remaining part of the circle are in
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