current attempt in progress\nuse the figure below to estimate ∫-10,15 f(x) dx.\n∫-10,15 f(x)dx = \netextbook…

Explanation:

Step1: Divide the area under - curve

We can divide the region between the curve $y = f(x)$, the $x$-axis, $x=-10$ and $x = 15$ into simple geometric shapes (approximate rectangles and triangles).

Step2: Estimate the area of each part

From $x=-10$ to $x = 0$: We can approximate this part as a trapezoid. The bases of the trapezoid are $b_1\approx0$ and $b_2\approx15$, and the height $h = 10$. The area of a trapezoid $A_1=\frac{(b_1 + b_2)h}{2}=\frac{(0 + 15)\times10}{2}=75$. From $x = 0$ to $x=10$: We can approximate this part as a rectangle - like shape with an average height of about $h_2\approx12$ and width $w_2 = 10$. The area $A_2=12\times10 = 120$. From $x = 10$ to $x = 15$: We can approximate this part as a triangle. The base of the triangle $b_3=5$ and the height $h_3\approx20$. The area of a triangle $A_3=\frac{1}{2}\times b_3\times h_3=\frac{1}{2}\times5\times20 = 50$.

Step3: Sum up the areas

The definite - integral $\int_{-10}^{15}f(x)dx$ is approximately the sum of the areas of these shapes. So, $\int_{-10}^{15}f(x)dx\approx A_1+A_2+A_3$. $\int_{-10}^{15}f(x)dx\approx75 + 120+50=245$.

Answer:

245