# Find area of the region under the curve y = 5 x^3 - 9 and above the x-axis, for 3 less than or...

## Question:

Find area of the region under the curve {eq}y = 5 x^3 - 9 {/eq} and above the x-axis, for {eq}\displaystyle 3 \le x \le 6 {/eq}.

## The Definite Integral as Area Under a Curve:

If {eq}f(x) {/eq} is a continuous function for {eq}b \geq a {/eq}, the area of the region bounded by the function and {eq}x-axis {/eq} on the interval {eq}\left[a,b\right] {/eq} is

{eq}Area=\int_{a}^{b}f(x)dx {/eq}

Step 1. Set up the definite integral that is equivalent to the area under the curve and above the x-axis.

{eq}\begin{align} \displaystyle Area&=\int_{a}^{b}f(x)dx\\ &=\int_{3}^{6}\left(5x^{3}-9\right)\:dx\\ \\ \end{align} {/eq}

Step 2. Evaluate the integral.

{eq}\begin{align} \displaystyle \int_{3}^{6}\left(5x^{3}-9\right)\:dx&=\left[\frac{5}{4}x^{4}-9x\right]_{x=3}^{x=6}\\ &=\frac{5}{4}(6)^{4}-9(6)-\frac{5}{4}(3)^{4}+9(3)\\ &=\frac{5967}{4}\\ \\ \end{align} {/eq}

Therefore the area under the curve and above the x-axis is {eq}\displaystyle \frac{5967}{4} {/eq}.