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3 years ago

Evaluate the integral by reversing the order of integration. Integral from 0 to 1 integral from √x to 1 of 5/(y³+1) dydx

Mathematics

1 answer:

ELEN [110]3 years ago

7 0

Answer:

⁵/₃ ln 2

Step-by-step explanation:

∫₀¹ ∫√ₓ¹ 5 / (y³ + 1) dy dx

We want to change the order of integration. To do this, we start by graphing the region contained by the four limits.

x = 0, x = 1

y = √x, y = 1

(Notice that y = √x is the same as x = y²).

Once we've graphed the region, we need to write the domain of x in terms of y. In this case, 0 < x < y².

Then, we find the range: 0 < y < 1.

Now we can rewrite the integral:

∫₀¹ ∫₀ʸ² 5 / (y³ + 1) dx dy

Notice that the integrand itself, 5 / (y³ + 1), does not change. Only the limits have changed.

Solve the integral.

∫₀¹ [ 5 / (y³ + 1) x |₀ʸ² ] dy

∫₀¹ [ 5y² / (y³ + 1) ] dy

⁵/₃ ∫₀¹ [ 3y² / (y³ + 1) ] dy

(⁵/₃ ln|y³ + 1|) |₀¹

⁵/₃ ln|1³ + 1| − ⁵/₃ ln|0³ + 1|

⁵/₃ ln 2

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2³+2²+2÷2²+2+1=2 prove that this relation is true for any natural numbers

UkoKoshka [18]

So, this question is basically asking us "If we had an x instead of a 2, would this be true?" We can try and see what we get:

$\frac{x^3+x^2+x}{x^2+x+1}=x$

So, if we want to show this we have to change the numerator or denominator in such a way that we can cancel some common factors. Notice that $x^3+x^2+x=x(x^2+x+1)$

If we replace the factored numerator with the original one, we get:

$\frac{x(x^2+x+1)}{x^2+x+1}=x \implies \\ x=x$

Since we have an equality, this relation is proved.

3 0

3 years ago

Final exam scores are normally distributed with a mean of 74 and a standard deviation of 6. Approximately, what percentage of fi

notka56 [123]

Answer:

81.86%

Step-by-step explanation:

We have been given that final exam scores are normally distributed with a mean of 74 and a standard deviation of 6.

First of all we will find z-score using z-score formula.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{68-74}{6}$

$z=\frac{-6}{6}=-1$

Now let us find z-score for 86.

$z=\frac{86-74}{6}$

$z=\frac{12}{6}=2$

Now we will find P(-1<Z) which is probability that a random score would be greater than 68. We will find P(2>Z) which is probability that a random score would be less than 86.

Using normal distribution table we will get,

$P(-1$