Question 369761
Assuming the equation is
{{{y = 4x + (1/x^2)}}} (Please use "^" to indicate exponents.)<br>
The answer will the the value of the definite integral:
{{{int(4x + 1/x^2, dx, 1, 4)}}}
Since integrals of powers of x are simple, I am going to rewrite the second term as the appropriate power of x:
{{{int(4x + x^(-2), dx, 1, 4)}}}
Next we can use a property of integrals that tells us the the integral of a sum is the same as the sum of the integrals:
{{{int(4x, dx, 1, 4) + int(x^(-2), dx, 1, 4)}}}
Next we can factor out the 4 in the first integral using a property of integrals:
{{{4*int(x, dx, 1, 4) + int(x^(-2), dx, 1, 4)}}}
Each integral is now a power of x. In general {{{int(x^n, dx) = x^(n+1)/(n+1)}}}. Using this on both of our integrals we get:
{{{4(x^(1+1)/(1+1)) + x^(-2+1)/(-2+1)}}} evaluated from x=1 to x=4
This simplifies as follows:
{{{4(x^2/2) + x^(-1)/(-1)}}} evaluated from x=1 to x=4
{{{2x^2 + (-x^(-1))}}} evaluated from x=1 to x=4
Evaluating this expression from 1 to 4:
{{{(2(4)^2 + (-(4)^(-1))) - (2(1)^2 + (-(1)^(-1)))}}}
Simplifying:
{{{(2(16) + (-1/4)) - (2*1 + (-1))}}}
{{{(32 + (-1/4)) - (2 + (-1))}}}
{{{(128/4 + (-1/4)) - (1)}}}
{{{(127/4) - (1)}}}
{{{(127/4) - (4/4)}}}
{{{123/4}}}