find the total area bounded by the x - axis and the curve y = f(x) on the indicated interval. f(x)=x² + 4x…

Explanation:

Step1: Recall area - integral formula

The area $A$ between the curve $y = f(x)$ and the $x$-axis on the interval $[a,b]$ is given by $A=\int_{a}^{b}|f(x)|dx$. First, we need to check the sign of $y = f(x)=x^{2}+4x - 5$ on the interval $[0,6]$. We find the roots of $f(x)$ by setting $f(x)=0$, so $x^{2}+4x - 5=(x + 5)(x - 1)=0$. The roots are $x=-5$ and $x = 1$. On the interval $[0,6]$, when $x\in[0,6]$, $f(x)=x^{2}+4x - 5$ is non - negative. So, $A=\int_{0}^{6}(x^{2}+4x - 5)dx$.

Step2: Apply integral rules

We know that $\int(x^{2}+4x - 5)dx=\int x^{2}dx+4\int xdx-\int 5dx$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int x^{2}dx=\frac{x^{3}}{3}$, $\int xdx=\frac{x^{2}}{2}$, and $\int 5dx=5x$. So, $\int(x^{2}+4x - 5)dx=\frac{x^{3}}{3}+4\times\frac{x^{2}}{2}-5x=\frac{x^{3}}{3}+2x^{2}-5x$.

Step3: Evaluate the definite integral

$A=\left[\frac{x^{3}}{3}+2x^{2}-5x\right]_{0}^{6}$. Substitute the upper and lower limits: $A=\left(\frac{6^{3}}{3}+2\times6^{2}-5\times6\right)-\left(\frac{0^{3}}{3}+2\times0^{2}-5\times0\right)$. Calculate $\frac{6^{3}}{3}=\frac{216}{3}=72$, $2\times6^{2}=2\times36 = 72$, and $5\times6 = 30$. Then $A=(72 + 72-30)-0=114$.

Answer:

$114$