find the total area between the graph of the function f(x) = x + 1 and the x - axis over the interval -2,5…

Explanation:

Step1: Find where the function crosses x - axis

Set $f(x)=x + 1=0$, then $x=-1$.

Step2: Split the integral based on the crossing - point

We split the interval $[-2,5]$ into $[-2,-1]$ and $[-1,5]$. The area $A=\left|\int_{-2}^{-1}(x + 1)dx\right|+\int_{-1}^{5}(x + 1)dx$.

Step3: Calculate $\int_{-2}^{-1}(x + 1)dx$

Using the power - rule $\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int_{-2}^{-1}(x + 1)dx=\left[\frac{x^{2}}{2}+x\right]{-2}^{-1}=\left(\frac{(-1)^{2}}{2}-1\right)-\left(\frac{(-2)^{2}}{2}-2\right)=\left(\frac{1}{2}-1\right)-(2 - 2)=-\frac{1}{2}$. So, $\left|\int{-2}^{-1}(x + 1)dx\right|=\frac{1}{2}$.

Step4: Calculate $\int_{-1}^{5}(x + 1)dx$

$\int_{-1}^{5}(x + 1)dx=\left[\frac{x^{2}}{2}+x\right]_{-1}^{5}=\left(\frac{5^{2}}{2}+5\right)-\left(\frac{(-1)^{2}}{2}-1\right)=\left(\frac{25}{2}+5\right)-\left(\frac{1}{2}-1\right)=\frac{25 + 10}{2}-\frac{1 - 2}{2}=\frac{35}{2}+\frac{1}{2}=18$.

Step5: Calculate the total area

$A=\frac{1}{2}+18=\frac{1 + 36}{2}=\frac{37}{2}$.

Answer:

$\frac{37}{2}$