If f(x) = \int_0^{x^4}\sqrt{t^2 + 4}dt then find f'(x). Hint : You will need to use the chain rule.

Question:

If {eq}f(x) = \int_0^{x^4}\sqrt{t^2 + 4}dt {/eq} then find f'(x).

Hint : You will need to use the chain rule.

Definite Integrals:

A definite integral would be the quantity at which the value of a function is accumulated over a certain range within its domain. We would acquire the definite integral through the various methods, and one way to solve this is through an analytical approach, wherein we apply the fundamental theorem of calculus.

Answer and Explanation: 1

Differentiate the given expression. We do this by first applying the fundamental theorem of calculus and use the formula for integrating functions as

{eq}\displaystyle \int_a^b g(t) dt = G(b) - G(a) {/eq}

where g(t) is the function for the integrand and G(t) is its antiderivative expression. For this problem, we assign

{eq}\displaystyle g(t) = \sqrt{t^2+4} {/eq}

and have its antiderivative expression to be G(t), wherein

{eq}\displaystyle G'(t) = g(t) {/eq}

Now, we apply the formula over the given integral and then evaluate its derivative to find f'(x). We would need to apply the chain rule, which is expressed as

{eq}\displaystyle \left( f(g(x)) \right)' = f'(g(x))g'(x) {/eq}

We proceed with the solution.

{eq}\begin{align} \displaystyle f(x) &= \int_0^{x^4}\sqrt{t^2 + 4}dt \\[0.3cm] &= \int_0^{x^4}g(t)dt \\[0.3cm] &= G(x^4) - G(0) \\[0.3cm] f'(x) &= \left( G(x^4) - G(0) \right)' \\[0.3cm] &= G'(x^4)\cdot 4x^3 - 0 \\[0.3cm] &= 4x^3\ g(x^4)\\[0.3cm] &= 4x^3 \sqrt{(x^4)^2 + 4} \\[0.3cm] &= \boxed{4x^3\sqrt{ x^8+4 }} \end{align} {/eq}


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Evaluating Definite Integrals Using the Fundamental Theorem

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Chapter 16 / Lesson 2
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The fundamental theorem of calculus makes finding your definite integral almost a piece of cake. See how the definite integral becomes a subtraction problem after applying the fundamental theorem of calculus.


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