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Find \frac{d}{dx}\int_{1}^{4x} \sqrt{t^2 + 1}\ dt.

Question:

Find {eq}\frac{d}{dx}\int_{1}^{4x} \sqrt{t^2 + 1}\ dt. {/eq}

Fundamental Theorem of Calculus

Finding the derivative of a function defined as an integral may seem to be a difficult task on the surface. However, there's a shortcut outlined in the Fundamental Theorem of Calculus that we can use to find this derivative quickly.

{eq}\frac{d}{dx} \int_a^x f(t) dt = f(x)\\ \frac{d}{dx} \int_{g(x)}^{h(x)} f(t) dt = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x) {/eq}

Answer and Explanation:

Rather than try to evaluate this integral, we can jump right to finding the derivative using the Fundamental Theorem of Calculus. Since the upper bound of this integral is a function of x, not simply x, we need to combine this theorem with the Chain Rule. We can ignore the lower bound, as it's a constant. This means that we can find this derivative using the procedure below.

{eq}\begin{align*} \frac{d}{dx}\int_{1}^{4x} \sqrt{t^2 + 1}\ dt &= \sqrt{(4x)^2+1} \cdot \frac{d}{dx} 4x\\ &= \sqrt{16x^2+1} \cdot 4\\ &= 4 \sqrt{16x^2 + 1} \end{align*} {/eq}


Learn more about this topic:

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The Fundamental Theorem of Calculus

from Math 104: Calculus

Chapter 12 / Lesson 10
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