Derivatives of integrals Simplify the followi...

Question

Derivatives of integrals Simplify the following expressions.

d/dt ∫₀ᵗ d𝓍/(1 + 𝓍²) + ∫₁¹/ᵗ dx/(1 + 𝓍²)

Accepted Answer

Welcome back, everyone. In this problem, we want to evaluate and simplify the derivative with respect to T of the integral of DX divided by 1 + X2 between the limits of -1 and T, plus the derivative of DX divided by 1 + X2 between the limits 3 and 1 divided by T. A says it's 1 divided by 1 + T2, B 2 divided by 1 + T2, C 0, and D times T divided by 1 + T squared. Now to help us evaluate this integral, then we can use or what would be helpful is the fundamental theorem of calculus, OK? I know if we recall that theorem. Mhm. It basically tells us. That if an antiderivative of F of X is equal to the integral of FFT with respect to T between the limits A and X, OK. Then the derivative. Of F of X. Or the derivative of the antiderivative rather of FF X will be equal to FFX, provided that FFX is continuous. So that's gonna be pretty helpful in helping us to figure out the derivative of our antiderivatives here. So let's break them up into two terms. Let's first start by differentiating the first integral. Now if we apply or if we use that theorem, then this tells us that the derivative with respect to t of the integral of D X divided by 1 + X2 between -1 and T is going to be equal to 1 divided by 1 + T2. OK, so that's our derivative for our first term. Now to differentiate or a second term, then we can recall the Leibniz rule, which is basically the fundamental theorem of calculus with the chain rule and. And if we recall the Leibniz rule, OK. Can live next. Then it basically tells us that the derivative with respect to T of the integral of F of X with respect to X between A and or function G of T. Is going to be equal to F of G of T multiplied by the derivative of G of T. So how can we apply that to our problem? Well, let's let F of X can let F of X here. Be equal to 1 divided by 1 + X2. Well, OK, if that's the case, then G of T is gonna be equal to 1 divided by T, which we can tell here from our limit in our second term, OK? And if G of T equals 1 divided by T, then this implies that the derivative of G of T is going to be equal to -1 divided by T2. So, since we have G of T and its derivative and F of X, Then that means our derivative with respect to T of the integral of D X divided by 1 + X2 between 3 and 1 divided by T is going to be equal to 1 divided by 1 + 1 divided by T2, OK, multiplied. By the derivative of G of T, which is -1 divided by T2. If we combine both of those, then this is gonna be -1 divided by T2 multiplied by 1 plus. Here one Plus 1 divided by T2 and if we simplify our denominator, we end up with -1 divided by T2 + 1. Now that we have our derivative for our second term, we can add both derivatives together. Because no, this implies that it's going to be equal to our first derivative 1 divided by 1 + T2 plus our second derivative, which is -1 divided by T2 + 1 or we could rewrite as 1 + T2. And when we do that, you notice here this is really saying that we're subtracting the same term from itself, and we know that that is going to be equal to zero. Thus, this tells us that our simplified version of our integral, when we simplify it, it will be equal to 0. If we look back at our answer choices, that means C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Derivatives of integrals Simplify the following expressions.

d/dt ∫₀ᵗ d𝓍/(1 + 𝓍²) + ∫₁¹/ᵗ dx/(1 + 𝓍²)

Key Concepts

Fundamental Theorem of Calculus

Chain Rule

Improper Integrals

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