Find the value of x. 2(x-3) x 8(x+2) = 128

Mathematics
Find the value of x. 2(x-3) x 8(x+2) = 128

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Answer

24

Step 1: ∫152x dx\int_1^5 2x \, dx

∫2x dx=x2+C\int 2x \, dx = x^2 + C [x2]15=52−12\left[ x^2 \right]_1^5 = 5^2 - 1^2 52=25,12=15^2 = 25, \quad 1^2 = 1 25−1=2425 - 1 = 24

∫152x dx=24\int_1^5 2x \, dx = 24

24

Step 2: ∫15(2x+3) dx\int_1^5 (2x + 3) \, dx

∫(2x+3) dx=x2+3x+C\int (2x + 3) \, dx = x^2 + 3x + C [x2+3x]15=(52+3⋅5)−(12+3⋅1)\left[ x^2 + 3x \right]_1^5 = (5^2 + 3 \cdot 5) - (1^2 + 3 \cdot 1) 52+3⋅5=25+15=405^2 + 3 \cdot 5 = 25 + 15 = 40 12+3⋅1=1+3=41^2 + 3 \cdot 1 = 1 + 3 = 4 40−4=3640 - 4 = 36

∫15(2x+3) dx=36\int_1^5 (2x + 3) \, dx = 36

36

Step 3: f(x)=x2−3x+2f(x) = x^2 - 3x + 2, ∫15f(x) dx=∫15(x2−3x+2) dx\int_1^5 f(x) \, dx = \int_1^5 (x^2 - 3x + 2) \, dx

∫(x2−3x+2) dx=13x3−32x2+2x+C\int (x^2 - 3x + 2) \, dx = \frac{1}{3} x^3 - \frac{3}{2} x^2 + 2x + C [13x3−32x2+2x]15\left[ \frac{1}{3} x^3 - \frac{3}{2} x^2 + 2x \right]_1^5

At x=5x=5:

13(125)−32(25)+2(5)=1253−752+10\frac{1}{3} (125) - \frac{3}{2} (25) + 2(5) = \frac{125}{3} - \frac{75}{2} + 10

Common denominator 66:

1253=2506,752=2256,10=606\frac{125}{3} = \frac{250}{6}, \quad \frac{75}{2} = \frac{225}{6}, \quad 10 = \frac{60}{6} 2506−2256+606=250−225+606=856\frac{250}{6} - \frac{225}{6} + \frac{60}{6} = \frac{250 - 225 + 60}{6} = \frac{85}{6}

At x=1x=1:

13(1)−32(1)+2(1)=13−32+2\frac{1}{3} (1) - \frac{3}{2} (1) + 2(1) = \frac{1}{3} - \frac{3}{2} + 2

Common denominator 66:

13=26,32=96,2=126\frac{1}{3} = \frac{2}{6}, \quad \frac{3}{2} = \frac{9}{6}, \quad 2 = \frac{12}{6} 26−96+126=2−9+126=56\frac{2}{6} - \frac{9}{6} + \frac{12}{6} = \frac{2 - 9 + 12}{6} = \frac{5}{6} 856−56=806=403\frac{85}{6} - \frac{5}{6} = \frac{80}{6} = \frac{40}{3}

∫15(x2−3x+2) dx=403\int_1^5 (x^2 - 3x + 2) \, dx = \dfrac{40}{3}

\dfrac{40{3}}

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_1^5 2x \, dx 2x \, dx = x^2 + C [ x^2 ]_1^5 = 5^2 - 1^2 5^2 = 25, 1^2 = 1 25 - 1 = 24 _1^5 2x \, dx = 24 24 Step 2: _1^5 (2x + 3) \, dx (2x + 3) \, dx = x^2 + 3x + C [ x^2 + 3x ]_1^5 = (5^2 + 3 · 5) - (1^2 + 3 · 1) 5^2 + 3 · 5 = 25 + 15 = 40 1^2 + 3…

Find the value of x. 2(x-3) x 8(x+2) = 128
Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

ScanSolve AI Answer
Step 1: _1^5 2x \, dx 2x \, dx = x^2 + C [ x^2 ]_1^5 = 5^2 - 1^2 5^2 = 25, 1^2 = 1 25 - 1 = 24 _1^5 2x \, dx = 24 24 Step 2: _1^5 (2x + 3) \, dx (2x + 3) \, dx = x^2 + 3x + C [ x^2 + 3x ]_1^5 = (5^2 + 3 · 5) - (1^2 + 3 · 1) 5^2 + 3 · 5 = 25 + 15 = 40 1^2 + 3 · 1 = 1 + 3 = 4 40 - 4 = 36 _1^5 (2x + 3) \, dx = 36 36 Step 3: f(x) = x^2 - 3x + 2, _1^5 f(x) \, dx = _1^5 (x^2 - 3x + 2) \, dx (x^2 - 3x + 2) \, dx = (1)/(3) x^3 - (3)/(2) x^2 + 2x + C [ (1)/(3) x^3 - (3)/(2) x^2 + 2x ]_1^5 At x=5: (1)/(3) (125) - (3)/(2) (25) + 2(5) = (125)/(3) - (75)/(2) + 10 Common denominator 6: (125)/(3) = (250)/(6), (75)/(2) = (225)/(6), 10 = (60)/(6) (250)/(6) - (225)/(6) + (60)/(6) = (250 - 225 + 60)/(6) = (85)/(6) At x=1: (1)/(3) (1) - (3)/(2) (1) + 2(1) = (1)/(3) - (3)/(2) + 2 Common denominator 6: (1)/(3) = (2)/(6), (3)/(2) = (9)/(6), 2 = (12)/(6) (2)/(6) - (9)/(6) + (12)/(6) = (2 - 9 + 12)/(6) = (5)/(6) (85)/(6) - (5)/(6) = (80)/(6) = (40)/(3) _1^5 (x^2 - 3x + 2) \, dx = (40)/(3) (40)/(3)