For what positive numbers will the cube of a number exceeds four times its square

Mathematics

1 answer:

jekas [21]3 years ago

5 0

Starting from 4, all cubes of positive integers are more than 4 times their squares

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Integration of (3X(X^2+3)^4) dx<br><img src="https://tex.z-dn.net/?f=%20" id="TexFormula1" title=" " alt=" " align="absmiddle" c

dimaraw [331]

Answer:

Step-by-step explanation:

$\int 3x(x^2+3)^4 \ dx$ .

It is apparently obvious we could expand the bracket and integrate term-by-term. This method would work but is very time consuming (and you could easily make a mistake) so we use a different method: integration by substitution.

Integration by substitution involves swapping the variable $x$ for another variable which depends on x: $u(x)$ . (We are going to choose $u$ for this question).

The very first step is to choose a suitable substitution. That is, an equation $u=f(x)$ which is going to make the integration easier. There is a trick for spotting this however: if an integral contains both a term and it's derivative then use the substitution $u=\text{The Term}$ .

Your integral contains the term $x^2 + 3$ . The derivative is $2x$ and (ignoring the constants) we see $x$ is also in the integral and so the substitution $u=x^2+3$ will unravel this integral!

Step 2: We must now swap the variable of integration from x to u. That means interchanging all the x's in the integrand (the term being integrated) for u's and also swapping (dx" to "du").

$u=x^2+3 \Rightarrow \frac{du}{dx}=2x \Rightarrow dx = \frac{1}{2x} du$