Question

Evaluate the integral. 3 f(x) dx −4 where f(x) = 3 if −4 ≤ x ≤ 0 4 − x2 if 0 < x ≤ 3

Answer 1

It looks like you have

$f(x) = \begin{cases} 3 & \text{if } -4 \le x \le 0 \\ 4 - x^2 & \text{if } 0 < x \le 3\end{cases}$

and the integral you want to compute is

$\displaystyle \int_{-4}^3 f(x) \, dx$

Split the integral up at $x=0$. Then

$\displaystyle \int_{-4}^3 f(x) \, dx = \int_{-4}^0 f(x)\,dx + \int_0^3 f(x)\,dx \\\\ ~~~~~~~~~~~~ = \int_{-4}^0 3 \, dx + \int_0^3 (4 - x^2) \, dx \\\\ ~~~~~~~~~~~~ = 3(0 - (-4)) + \left(4\cdot3 - \frac{3^3}3\right) = \boxed{15}$

Answer 2

After evaluating the integral we get 15

Integral is the representation of the area of a region under a curve. We approximate the actual value of an integral by drawing rectangles. A definite integral of a function can be represented as the area of the region bounded by its graph of the given function between two points in the line.

Given,

f(x) = 3         if   - 4 ≤ x ≤ 0

f(x) = 4 - $x^{2}$  if   0 < x ≤ 3

Then,

$f(x) = \left \{ {{3} \atop {4-x^{2} }}$

We need to solve the integral -$\int\limits^3_4 {f(x)} \, dx$

Elaborate the integral with x = 0

                                             -$\int\limits^3_4 {f(x)} \, dx$

                                          = -$\int\limits^0_4 {f(x)} \, dx + \int\limits^3_0 {f(x)} \, dx$

                                          = -$\int\limits^0_4 {3} \, dx + \int\limits^3_0 {4-x^{2} } \, dx$

                                          = 3 ( 0 - (-4)) + (4.3 - $\frac{3^{3} }{3}$)

                                          = 15

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