What is #f(x) = int x^4-x^2+5x dx# if #f(3)=-1 #?

Question

What is #f(x) = int x^4-x^2+5x   dx# if #f(3)=-1  #?

Aurora Alvarado · Accepted Answer

#f(x)=x^5/5-x^3/3+(5x^2)/2-631/10#

Integrate each term individually using the rule:

#intx^ndx=(x^(n+1))/(n+1)+C#

Applying this, we see that

#f(x)=(x^(4+1))/(4+1)-(x^(2+1))/(2+1)+(5x^(1+1))/(1+1)+C#

#f(x)=x^5/5-x^3/3+(5x^2)/2+C#

Since we know that #f(3)=-1#, we can find #C#, the constant of integration.

#f(3)=(3^5)/5-3^3/3+(5(3^2))/2+C=-1#

#243/5-9+45/2+C=-1#

#486/10-90/10+225/10+C=-10/10#

#C=-631/10#

Thus,

#f(x)=x^5/5-x^3/3+(5x^2)/2-631/10#

Charles Anctil · Answer

undefined To find $ f(x) = \int (x^4 - x^2 + 5x) \, dx $ given that $ f(3) = -1 $, integrate the function with respect to $ x $ and then use the given condition to solve for the constant of integration.

$$ f(x) = \int (x^4 - x^2 + 5x) \, dx $$

$$ f(x) = \frac{1}{5}x^5 - \frac{1}{3}x^3 + \frac{5}{2}x^2 + C $$

Given that $ f(3) = -1 $, substitute $ x = 3 $ into the equation:

$$ -1 = \frac{1}{5}(3)^5 - \frac{1}{3}(3)^3 + \frac{5}{2}(3)^2 + C $$

$$ -1 = \frac{243}{5} - 27 + \frac{45}{2} + C $$

$$ -1 = \frac{243}{5} - \frac{135}{5} + \frac{225}{10} + C $$

$$ -1 = \frac{243 - 135 + 112}{10} + C $$

$$ -1 = \frac{220}{10} + C $$

$$ -1 = 22 + C $$

$$ C = -1 - 22 $$

$$ C = -23 $$

Therefore, the function $ f(x) = \int (x^4 - x^2 + 5x) \, dx $ is:

$$ f(x) = \frac{1}{5}x^5 - \frac{1}{3}x^3 + \frac{5}{2}x^2 - 23 $$