Example 2: Definite Integral Power Rule

$\underset{3}{\overset{7}{\int }}{x}^3+{5x}^2–7x+3\mathit{ dx}$

Step 1: Simplify and look for algebraic rewrites.

When looking at the first chunk Remind yourself that there is really an unwritten 1 for the power. That is what allows us to apply the power rule .

$\underset{3}{\overset{7}{\int }}{x}^3+{5x}^2–7x^1+3\mathit{ dx}$

Step 2: Identify any term(s) that include variables raised to a power.

Break the problem down into bitesize chunks based upon the + and – , and identify the antiderivative rule for each chunk (term).

Here you have 4 chunks. The first three chunks will be a power rule , and the final chunk is a constant.

$\underset{3}{\overset{7}{\int }}{x}^3+{5x}^2–7x^1+3\mathit{ dx}$

Step 3: Take the antiderivativeof the variables raised to a power using the Recipe:Add 1 to the power; Divide by the new power.

Chunk 1: Power Rule

Chunk 2: Power Rule

Chunk 3: Power Rule

Chunk 4: Constant Rule

You might find it helpful (I do) to slide your fractions to the front of each term.

$\underset{3}{\overset{7}{\int }}{x}^3+{5x}^2–7x^1+3 \mathit{dx}$

$\mathit{=}{\frac{x^{3+1}}{4}+\frac{5x^{2+1}}{3}–\frac{7x^{1+1}}{2}+3x|}_{x=3}^{x=7}$

$\mathit{=}{\frac{x^4}{4}+\frac{5x^3}{3}–\frac{7x^2}{2}+3x|}_{x=3}^{x=7}$

$\mathit{=}{\frac{1}{4}{x}^4+\frac{5}{3}{x}^3–\frac{7}{2}{x}^2+3x|}_{x=3}^{x=7}$

Step 4 ( Definite Integral ONLY ): Evaluate the antiderivative result using the Top – Bottom method.

$\underset{3}{\overset{7}{\int }}{x}^3+{5x}^2–7x^1+3 \mathit{dx}$

$\mathit{=}{\frac{1}{4}{x}^4+\frac{5}{3}{x}^3–\frac{7}{2}{x}^2+3x|}_{x=3}^{x=7}$

$\mathit{=}\left(\frac{1}{4}{\left(7\right)}^4+\frac{5}{3}{\left(7\right)}^3–\frac{7}{2}{\left(7\right)}^2+3\left(7\right)\right)\mathit{–}\left(\frac{1}{4}{\left(3\right)}^4+\frac{5}{3}{\left(3\right)}^3–\frac{7}{2}{\left(3\right)}^2+3\left(3\right)\right)$

$\mathit{=}\left(\frac{12257}{12}\right)\mathit{–}\left(\frac{171}{4}\right)\mathit{=}\frac{2936}{3}$

Final Result Meaning: Remember the Definite Integral will always provide you a definite value , and the Indefinite Integral provides you a family of solutions .

The Net Area between the curve $f\left(x\right)=x^3+{5x}^2–7x+3$and the x-axis on the x-interval [ 3 , 7 ] is $\frac{2936}{3}$.

Since the final result is positive, you know without even seeing the graph that there is more area above the x-axis than below it.