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How do you evaluate the definite integral #int3xsin(x^2)# from #[0,sqrtpi]#?

How do you evaluate the definite integral #int3xsin(x^2)dx# from #[0,sqrtpi]#?

Answer 1

Paisley Johnson

#3.#

We know that, #intsin y dy=-cos y+C#.

Let #y=x^2," so that, "dy=2xdx#

#:. intsin x^2*2xdx=2intxsinx^2dx=-cosx^2+C#.

#:. intxsinx^2dx=-1/2cosx^2+C#

#rArr int3xsinx^2dx=3intxsinx^2dx=-3/2cosx^2+C#

Therefore,

#int_0^(sqrtpi) 3xsinx^2dx=-3/2[cosx^2]_0^(sqrtpi)#

#=-3/2[cospi-cos0]#

#=-3/2[-1-1]#

#=-3/2*(-2)#

#=3.#

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Answer 2

Ezekiel Alexander

To evaluate the definite integral ∫₃xsin(x²) from [0,√π], you can use the substitution method. Let u = x², then du/dx = 2x, or dx = du/(2x). The limits of integration will also change accordingly.

∫₃xsin(x²)dx = ∫₃√πsin(u) * (1/(2x)) du

Now, we need to find the new limits of integration when x = 0 and x = √π.

When x = 0, u = 0² = 0. When x = √π, u = (√π)² = π.

So, the new limits of integration are [0, π].

Now, rewrite the integral in terms of u:

∫₃xsin(x²)dx = ∫₀πsin(u) * (1/(2x)) du

Apply the integration:

= (-1/2) ∫₀πsin(u) du

= (-1/2) [-cos(u)] from 0 to π

= (-1/2) * [cos(π) - cos(0)]

= (-1/2) * [-1 - 1]

= 1

Therefore, the value of the definite integral ∫₃xsin(x²)dx from [0,√π] is 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.