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Обчисли площу фігури, обмеженої лініями y=(1)/(3^(x)),x-3y+3=0,x=3". "

Обчисли площу фігури, обмеженої лініями\newliney=13x,x−3y+3=0,x=3.  y=\frac{1}{3^{x}}, x-3 y+3=0, x=3 \text {. }

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Q. Обчисли площу фігури, обмеженої лініями\newliney=13x,x−3y+3=0,x=3.  y=\frac{1}{3^{x}}, x-3 y+3=0, x=3 \text {. }
  1. Rewrite Equation: Rewrite the equation x−3y+3=0x - 3y + 3 = 0 to express yy in terms of xx.\newliney=x+33y = \frac{x + 3}{3}
  2. Set Range for xx: Set the range for xx based on the given x=3x = 3.\newlineThe range of xx is from negative infinity to 33.
  3. Set up Integral: Set up the integral to find the area between y=13xy = \frac{1}{3^x} and y=x+33y = \frac{x + 3}{3} from x=−∞x = -\infty to x=3x = 3.\newlineArea = ∫−∞3[x+33−13x]dx\int_{-\infty}^{3} \left[\frac{x + 3}{3} - \frac{1}{3^x}\right] dx

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