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3 years ago

Suppose the function F is defined by F(x)= integral from sqrt x to 1 of (2t - 1)/(t + 2) dt for all real numbers x > or = 0?

1 answer:

7 0

F(x)= ∫[√x, 1] (2t - 1)/(t + 2) dt for x ≥ 0

Let f(x) = ∫(2t - 1)/(t + 2) dt = ∫[2 + -5/(t + 2)] dt = 2t - 5Ln(t + 2) + constant 

F(x) = f(1) - f(√x) = 2 – 5Ln(3) - (2√x - 4Ln(√x + 2) 

F(1) = 2 - 5Ln(3) - (2√1 - 5Ln(√1 + 2) = 2 - 5Ln(3) - (2 - 5Ln(3)) = 0

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3 years ago

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3 years ago

Find the length of the shorter leg of a right triangle with the longer leg is 9 feet more than the shorter leg the hypotenuse is

Jobisdone [24]

Answer:

Length of Shorter leg = 27 feet

Length of Longer leg = 36 feet

Length of Hypotenuse = 45 feet

Step-by-step explanation:

Given:

Let,

Length of Shorter leg = x,

According to the given condition we have

∴ Length of Longer leg = x + 9 feet

∴ Length of Hypotenuse = 2x - 9 feet

To Find:

Length of Shorter leg = ?

Solution:

Now In Right Angle Triangle, By Pythagoras theorem we have

$(\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}$

substituting the given values we get

$(2x-9)^{2} = (x)^{2}+(x+9)^{2}\\$

By using Identity (A ± B)² = A² ± 2AB +B² we get

$4x^{2} -36x+81=x^{2} +(x^{2} +18x+81)\\\\4x^{2} -2x^{2} -36x-18x=81-81\\\\2x^{2} -54x=0\\2x(x-27)=0\\2x=0 or x-27=0\\\therefore x\ cannot\ be \ zero\\\therefore x = 27 \feet$

Substituting x=27 we get

∴ Length of Longer leg = x + 9 =27 +9 = 36 feet

∴ Length of Hypotenuse = 2x - 9 = 2×27 -9 = 45 feet

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3 years ago

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