Question #946b3

2 Answers
Jan 28, 2017

int (x^{1/2} + 3x + 5) dx = 2/3 x^{3/2} + 3/2 x^2 + 5x + "constant"

Explanation:

Use the power rule.

For n != -1,

int x^n dx = x^{n+1}/{n+1} + "constant".

Therefore,

int (x^{1/2} + 3x + 5) dx = int x^{1/2} dx + 3int x dx + 5int x^0 dx

= 2/3 x^{3/2} + 3/2 x^2 + 5x + "constant"

Jan 28, 2017

2/3x^(3/2)+3/2x^2+5x+c

Explanation:

This is a Calculus question.

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

color(red)(bar(ul(|color(white)(2/2)color(black)(intax^ndx=a/(n+1)x^(n+1) ; n≠-1)color(white)(2/2)|)))

color(blue)"Note that " 5=5x^0" since " x^0=1

rArrint(x^(1/2)+3x+5)dx

=1/(3/2)x^((1/2+1))+3/2x^((1+1))+5x^((0+1))+c

=2/3x^(3/2)+3/2x^2+5x+c

where c is the constant of integration.