Use the second fundamental theorem of calculus to calculate F(x)=int_-3^x(t^2+3t+2)dt?

Sep 24, 2017

$F \left(x\right) = \frac{{x}^{3}}{3} + \frac{3}{2} {x}^{2} + 2 x$

Explanation:

The second fundamental theorem of calculus states that for some function $f \left(x\right)$ that is continuous over the interval $\left[a , b\right]$ (where $a$ is a constant), there exists a function $F \left(x\right) = {\int}_{a}^{x} f \left(t\right) d t$.

This means that $F \left(x\right)$ is an antiderivative of $f \left(x\right)$, or $F ' \left(x\right) = f \left(x\right) ,$ for all $x$ in $b$.

$R i g h t a r r o w \frac{d}{\mathrm{dx}} \left({\int}_{a}^{x} f \left(t\right) d t\right) = f \left(x\right)$

Let's substitute our function into this form:

$R i g h t a r r o w \frac{d}{\mathrm{dx}} \left({\int}_{- 3}^{x} \left({t}^{2} + 3 t + 2\right) d t\right) = {x}^{2} + 3 x + 2$

$\therefore F ' \left(x\right) = {x}^{2} + 3 x + 2$

Now that we have $F ' \left(x\right)$, we can evaluate its antiderivative, which will give us $F \left(x\right)$.

First, let's evaluate the indefinite integral of $F ' \left(x\right)$:

$R i g h t a r r o w \int \left({x}^{2} + 3 x + 2\right) d x = \int \left({x}^{2}\right) d x + \int \left(3 x\right) d x + \int \left(2\right) d x$

$R i g h t a r r o w \int \left({x}^{2} + 3 x + 2\right) d x = \frac{{x}^{2 + 1}}{2 + 1} + \frac{3 {x}^{1 + 1}}{1 + 1} + 2 x + C$

$\therefore \int \left({x}^{2} + 3 x + 2\right) d x = \frac{{x}^{3}}{3} + \frac{3 {x}^{2}}{2} + 2 x + C$

Then, let's calculate the definite integral in the interval $\left[- 3 , x\right]$:

$R i g h t a r r o w {\int}_{- 3}^{x} \left({x}^{2} + 3 x + 2\right) d x = F \left(x\right) - F \left(- 3\right)$

Now, $F \left(- 3\right) = {\int}_{- 3}^{- 3} \left({x}^{2} + x + 2\right) d x = 0$, because the integration is being done over an interval of length $0$:

$R i g h t a r r o w {\int}_{- 3}^{x} \left({x}^{2} + 3 x + 2\right) d x = \left(\frac{{x}^{3}}{3} + \frac{3 {x}^{2}}{2} + 2 x\right) - \left(0\right)$

$R i g h t a r r o w {\int}_{- 3}^{x} \left({x}^{2} + 3 x + 2\right) d x = \frac{{x}^{3}}{3} + \frac{3 {x}^{2}}{2} + 2 x$

$R i g h t a r r o w {\int}_{- 3}^{x} \left({x}^{2} + 3 x + 2\right) d x = \frac{{x}^{3}}{3} + \frac{3}{2} {x}^{2} + 2 x$

$\therefore F \left(x\right) = \frac{{x}^{3}}{3} + \frac{3}{2} {x}^{2} + 2 x$