Test your Integration Skills 103!

Calculus Level 4

If $\displaystyle f(x) = \int_2^{2x} \dfrac1{\sqrt{t^3+1}} \, dt$ , find $f'(1)$ .

\sqrt2

\frac23

0

\frac13

2 solutions

Indre Dan
Mar 28, 2016

A possible solution to avoid the integration is using the following formula:
Let be the function $f:D\rightarrow \mathbb{R} \, \; f(x)=\int_{u(x)}^{v(x)}F(t)dt$
Where: $f'(x)=v'(x)F(v(x))-u'(x)F(u(x))$

Going back to our integral we find that:

$f'(x)=(\int_{2}^{2x}\frac{1}{\sqrt{t^3+1}}dt)'=2x'\frac{1}{\sqrt{(2x)^3+1}}=\frac{2}{\sqrt{8x^3+1}}$ $f'(1)=\frac{2}{\sqrt{8\ast1+1 }}=\frac{2}{\sqrt{9}}=\frac{2}{3}$

The result here is a pretty important one, and is known as the Newton-Leibnitz rule for differentiation under integral sign. (+1)

Pulkit Gupta - 5 years, 2 months ago

Sanjoy Kundu
Jan 5, 2019

Trivial by the Fundamental Theorem of Calculus, how is this a level 4 problem...

Test your Integration Skills 103!

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