How do you evaluate #int (1 + x absx ) dx# for [-2, 3]?

Question

Evelyn Agard · Accepted Answer

#34/3# #|x| = x# for #AAx>=0# #|x| = -x# for #AAx<0# #I = int_-2^3 (1+x|x|)dx = int_-2^0 (1+x(-x))dx + int_0^3 (1+x*x)dx# #I = int_-2^0 (1-x^2)dx + int_0^3 (1+x^2)dx# #I_1 = (x-x^3/3) = 0-0 - (-2) + (-2)^3/3 = 2-8/3 = -2/3# #I_2 = (x+x^3/3) = 3+3^3/3-0-0=12# #I = -2/3 + 12 = 34/3#

Axel Alvarez · Answer

undefined To evaluate the integral $\int (1 + x \left| x \right|) \, dx$ over the interval $[-2, 3]$, you first need to break it into two separate integrals based on the sign of $x$, as the absolute value function changes behavior at $x = 0$.

1. For $x < 0$: 
$\int_{-2}^{0} (1 + x(-x)) \, dx$
$$ = \int_{-2}^{0} (1 - x^2) \, dx$$

2. For $x \geq 0$:
$\int_{0}^{3} (1 + x \cdot x) \, dx$
$$ = \int_{0}^{3} (1 + x^2) \, dx$$

Now, integrate each part separately:

1. For $x < 0$: 
$$= \left[ x - \frac{x^3}{3} \right]_{-2}^{0}$$

2. For $x \geq 0$:
$$= \left[ x + \frac{x^3}{3} \right]_{0}^{3}$$

Evaluate each expression at the upper and lower limits of integration and then subtract the result of the lower limit from the upper limit for each part.

After performing the calculations, you will obtain the value of the integral over the given interval $[-2, 3]$.

Axel Alvarez · Answer

undefined To evaluate ∫(1 + x|𝑥|)𝑑𝑥 from -2 to 3, first split the integral into two parts: ∫(1 + x𝑥)𝑑𝑥 from -2 to 0 and ∫(1 + x𝑥)𝑑𝑥 from 0 to 3. Then integrate each part separately, and sum the results. For the first part (-2 to 0), integrate (1 + x𝑥) with respect to 𝑥, and for the second part (0 to 3), integrate (1 + x𝑥) with respect to 𝑥. Finally, subtract the result of the first part from the result of the second part to get the total value of the integral.