find derivative of integral( t sint dt) upper limit 1+2x, lower limit 1-2x
Show step by step please!
Thank you
Well, this is going to be a fun one! To find the derivative of the given integral, we can use the Leibniz rule for differentiating under the integral sign. So let's get started step by step!
First, let's rewrite the integral using integration by parts. We will let u = t and dv = sint dt. This means du = dt and v = -cost.
Now we have the integral of -t cost dt. But don't worry, we're not done yet!
Next, we'll apply the limits of integration. Let's substitute the upper and lower limits into the integral. We have:
integral from 1-2x to 1+2x of (-t cost) dt.
Now, let's plug in the upper limit first. We get:
[(-t cost) evaluated from 1-2x to 1+2x]
Next, we'll evaluate the integral at the lower limit. We have:
(-1-2x)(cos(1+2x))
Now, let's subtract the result of evaluating the integral at the lower limit from the result of evaluating the integral at the upper limit. We get:
[(-1-2x)(cos(1+2x)) - (-1+2x)(cos(1-2x))]
And finally, we differentiate this expression with respect to x. Applying the chain rule and product rule, we get:
d/dx [(-1-2x)(cos(1+2x)) - (-1+2x)(cos(1-2x))]
= -2cos(1+2x) + (1+2x)(-2sin(1+2x))(2)
-2cos(1-2x) + (1-2x)(2sin(1-2x))(-2)
And that's it! That's the derivative of the given integral with respect to x. I hope you enjoyed this step-by-step journey through the land of derivatives and integrals!
To find the derivative of the integral ∫(t * sin(t) dt) with respect to x, where the upper limit is 1 + 2x and the lower limit is 1 - 2x, we can use the Leibniz rule for differentiating under the integral sign.
Step 1: Apply the Leibniz rule:
d/dx ∫[a(x), b(x)] f(t) dt = f(b(x)) * d/dx(b(x)) - f(a(x)) * d/dx(a(x)) + ∫[a(x), b(x)] (d/dx)f(t) dt
In our case, f(t) = t * sin(t), a(x) = 1 - 2x, and b(x) = 1 + 2x.
Step 2: Calculate the derivatives of a(x) and b(x):
d/dx(a(x)) = d/dx(1 - 2x) = -2
d/dx(b(x)) = d/dx(1 + 2x) = 2
Step 3: Compute f(a(x)) and f(b(x)):
f(a(x)) = (1 - 2x) * sin(1 - 2x)
f(b(x)) = (1 + 2x) * sin(1 + 2x)
Step 4: Find the derivative of f(t) with respect to t:
(d/dt)(t * sin(t)) = sin(t) + t * cos(t)
Step 5: Substitute all the values into the Leibniz rule formula:
d/dx ∫[a(x), b(x)] f(t) dt = (1 + 2x) * sin(1 + 2x) * 2 - (1 - 2x) * sin(1 - 2x) * (-2) + ∫[a(x), b(x)] [(d/dt)(t * sin(t))] dt
Step 6: Simplify the expression:
2(1 + 2x) * sin(1 + 2x) + 2(1 - 2x) * sin(1 - 2x) + ∫[a(x), b(x)] [sin(t) + t * cos(t)] dt
And that's the derivative of the given integral!
To find the derivative of the integral \(\int_{1-2x}^{1+2x} t \sin(t) \, dt\), we can utilize the Fundamental Theorem of Calculus. The theorem states that if we have a function \(F(x)\) defined as an integral of a function \(f(t)\), i.e., \(F(x) = \int_{a}^{x} f(t) \, dt\), then the derivative of \(F(x)\) with respect to \(x\) is simply \(f(x)\).
So, in this case, we need to find the antiderivative of the integrand \(t \sin(t)\) and then evaluate it at the upper and lower limits to find the difference. Then we will take the derivative of this difference.
Step 1: Find the antiderivative of \(t \sin(t)\)
To find the antiderivative, we use integration by parts. The formula for integration by parts is \(\int u \, dv = uv - \int v \, du\).
Here, let \(u = t\) and \(dv = \sin(t) \, dt\).
\(\Rightarrow du = dt\) (since the derivative of \(t\) with respect to \(t\) is 1)
\(\Rightarrow v = -\cos(t)\) (antiderivative of \(\sin(t)\) is \(-\cos(t)\))
Using the integration by parts formula, we have:
\(\int t \sin(t) \, dt = -t \cos(t) - \int (-\cos(t) \, dt)\)
\(\Rightarrow \int t \sin(t) \, dt = -t \cos(t) + \int \cos(t) \, dt\)
Step 2: Evaluate the antiderivative at the limits
Now we need to evaluate the antiderivative at the upper and lower limits.
For the upper limit (let's call it \(x\)):
\(\int_{1-2x}^{1+2x} t \sin(t) \, dt = -t \cos(t) + \int \cos(t) \, dt \, \Big|_{1-2x}^{1+2x}\)
\(\Rightarrow = -(1+2x) \cos(1+2x) + \int \cos(1+2x) \, dt - [-(1-2x) \cos(1-2x) + \int \cos(1-2x) \, dt]\)
For the lower limit (let's call it \(a\)):
\(\Rightarrow = -(1+2a) \cos(1+2a) + \int \cos(1+2a) \, dt - [-(1-2a) \cos(1-2a) + \int \cos(1-2a) \, dt]\)
Step 3: Simplify and differentiate
Now, we can simplify this expression by evaluating the integrals and taking the derivative of the result.
Since \(\int \cos(t) \, dt = \sin(t) + C\), where \(C\) is the constant of integration, we can substitute this into our expression:
\(\Rightarrow = -(1+2x) \cos(1+2x) + (\sin(1+2x) + C) - [-(1-2x) \cos(1-2x) + (\sin(1-2x) + C)]\)
Now, differentiate this expression with respect to \(x\):
\(\frac{d}{dx} \left( - (1+2x) \cos(1+2x) + \sin(1+2x) - [-(1-2x) \cos(1-2x) + \sin(1-2x)] \right)\)
This will yield the derivative of the given integral.
Just use Leibniz's Rule:
d/dx(∫[1-2x,1+2x] t sint dt)
= (1+2x)sin(1+2x)(2) - (1-2x)sin(1-2x)(-2)
= 2(1+2x)sin(1+2x) + 2(1-2x)sin(1-2x)