How do you evaluate the definite integral #int (3-abs(x-3))dx# from [1,4]?

Question

Samantha Adkison · Accepted Answer

#int_1^4(3-abs(x-3))dx=13/2# The function #abs(x-3)# can be expressed as a piecewise function: #abs(x-3)={(x-3,,x>=3),(-(x-3),,x<3):}# So, we can write our integral as the sum of two integrals, breaking off at #x=3#: #int_1^4(3-abs(x-3))dx# #=int_1^3(3-(-(x-3)))dx+int_3^4(3-(x-3))dx# #=int_1^3(3+(x-3))dx+int_3^4(3-x+3)dx# #=int_1^3xcolor(white).dx+int_3^4(-x+6)dx# #=[x^2/2]_1^3+[-x^2/2+6x]_3^4# #=(3^2/2-1^2/2)+(-4^2/2+6(4))-(-3^2/2+3(6))# #=(9/2-1/2)+(-8+24)-(-9/2+18)# #=13/2#

Ezekiel Alexander · Answer

undefined To evaluate the definite integral $ \int_{1}^{4} (3 - |x - 3|) \, dx $, we split the integral into two parts based on the absolute value function.

1. For $ 1 \leq x \leq 3 $:
$$ 3 - |x - 3| = 3 - (3 - x) = x $$

2. For $ 3 < x \leq 4 $:
$$ 3 - |x - 3| = 3 - (x - 3) = 6 - x $$

Then, we integrate each part separately within the given limits:

1. For $ 1 \leq x \leq 3 $:
$$ \int_{1}^{3} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{3} = \frac{9}{2} - \frac{1}{2} = 4 $$

2. For $ 3 < x \leq 4 $:
$$ \int_{3}^{4} (6 - x) \, dx = \left[ 6x - \frac{x^2}{2} \right]_{3}^{4} = (24 - 8) - (18 - 4.5) = 16 - 13.5 = 2.5 $$

Finally, summing up the results from both intervals:
$$ \int_{1}^{4} (3 - |x - 3|) \, dx = 4 + 2.5 = 6.5 $$

William Abdalla · Answer

undefined To evaluate the definite integral ∫(3 - |x - 3|) dx from 1 to 4, you need to break the interval [1, 4] into two parts: from 1 to 3 and from 3 to 4. In each interval, the absolute value term changes sign at x = 3, so you'll have different expressions for the function in each part.

For the interval [1, 3], the function 3 - |x - 3| simplifies to 3 - (3 - x) = x.

For the interval [3, 4], the function 3 - |x - 3| simplifies to 3 - (x - 3) = 6 - x.

Now, integrate each part separately:
∫(x) dx from 1 to 3
= [x^2 / 2] evaluated from 1 to 3
= (9 / 2) - (1 / 2)
= 4

∫(6 - x) dx from 3 to 4
= [6x - (x^2 / 2)] evaluated from 3 to 4
= (24 - 8) - ((18 - 9) / 2)
= 16 - (9 / 2)
= 23 / 2

Add the integrals together:
4 + 23 / 2
= 8 + 23 / 2
= 31 / 2

So, the value of the definite integral ∫(3 - |x - 3|) dx from 1 to 4 is 31 / 2 or 15.5.