How do you find the antiderivative of #f(x)=1/4x^4-2/3x^2+4#?

Question

Isabella Ackerman · Accepted Answer

#intf(x)dx=int(1/4x^4-2/3x^2+4)dx#

#=1/4intx^4dx-2/3intx^2dx+4intx^0dx#

#=1/4{x^(4+1)/(4+1)}-2/3{x^(2+1)/(2+1)}+4{x^(0+1)/(0+1)}#

#=1/20x^5-2/9x^3+4x+C,#

# because, intx^ndx=x^(n+1)/(n+1)+c.#

Enjoy Maths.!

Emma Aldridge · Answer

#1/20x^5-2/9x^3+4x+c#

Integrate each term using the #color(blue)"power rule for integration"#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(int(ax^n)=a/(n+1)x^(n+1))color(white)(2/2)|)))#

#rArrint(1/4x^4-2/3x^2+4)dx#

#=(1/4)/5x^5-(2/3)/3x^3+4x+c#

#=1/20x^5-2/9x^3+4x+c#

where c is the constant of integration.

Axel Alvarez · Answer

undefined To find the antiderivative of $ f(x) = \frac{1}{4}x^4 - \frac{2}{3}x^2 + 4 $, integrate each term separately:
$$ \int \frac{1}{4}x^4 - \frac{2}{3}x^2 + 4 \, dx $$
$$ = \frac{1}{4} \int x^4 \, dx - \frac{2}{3} \int x^2 \, dx + \int 4 \, dx $$

Now, apply the power rule for integration to each term:
$$ = \frac{1}{4} \left( \frac{1}{5}x^5 \right) - \frac{2}{3} \left( \frac{1}{3}x^3 \right) + 4x + C $$

Combine the terms and add the constant of integration $ C $:
$$ = \frac{1}{20}x^5 - \frac{2}{9}x^3 + 4x + C $$

So, the antiderivative of $ f(x) = \frac{1}{4}x^4 - \frac{2}{3}x^2 + 4 $ is $ \frac{1}{20}x^5 - \frac{2}{9}x^3 + 4x + C $, where $ C $ is the constant of integration.