Question

How to solve it?1" title="\int\limits^a_b {x^2+2x} \, dx" alt="\int\limits^a_b {x^2+2x} \, dx" align="absmiddle" class="latex-formula">

Answer 1

Hi there! Assume that this is your question.

$\large{ \int \limits^a_b ( {x}^{2} + 2x)dx}$

Before we get to Integral, you have to know Differentiation first. If you know how to differentiate a polynomial function then we are good to go in Integral!

We call the function that we are going to integrate as Integrand. Integrand is a function that's differentiated. In Integral, Integrating requires you to turn the function from differentiated to an original function.

For Ex. If the Integrand is x² then the original function is (1/3)x³ because when we differentiate (1/3)x³, we get x²

$\large{f(x) = \frac{1}{3} {x}^{3} \longrightarrow f'(x) = {x}^{2} } \\ \large{f'(x) = 3( \frac{1}{3} ) {x}^{3 - 1} } \\ \large{f'(x) = {x}^{2} }$

So when we Integrate, make sure to convert Integrand as in original function. From the question, our Integrand is x²+2x. The function is in differentiated form. We know that x² is from (1/3)x³ and 2x comes from x²

$\large{ f(x) = {x}^{2} \longrightarrow f'(x) = 2x} \\ \large{f'(x) = 2 {x}^{2 - 1} } \\ \large{f'(x) = 2x}$

Thus,

$\large{ \int \limits^a_b ( {x}^{2} + 2x)dx} \\ \large{\int \limits^a_b ( \frac{1}{3} {x}^{3} + {x}^{2}) }$

Normally, if it's an indefinite Integral then we'd just put + C after (1/3)x³+x² but since we have a and b, it's a definite Integral.

$\large{ \int \limits^b_a f(x)dx = F(b) - F(a)}$

Define F(x) as our anti-diff

From our problem, substitute x = a in then subtract with the one that substitute x = b

$\large{ (\frac{1}{3}{a}^{3} + {a}^{2} ) - ( \frac{1}{3} {b}^{3} + {b}^{2}) }$

Simplify as we get:

$\large \boxed{ \frac{1}{3}{a}^{3} + {a}^{2} - \frac{1}{3} {b}^{3} - {b}^{2}}$