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find ∫01(f(x)−g(x)) dx\int_{0}^{1}(f(x)-g(x))\,dx, if ∫01(f(x)−2g(x)) dx=6\int_{0}^{1}(f(x)-2g(x))\,dx=6, ∫01(2f(x)+2g(x)) dx=9\int_{0}^{1}(2f(x)+2g(x))\,dx=9

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Evaluate definite integrals using the chain ruleright chevron icon

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Q. find ∫01(f(x)−g(x)) dx\int_{0}^{1}(f(x)-g(x))\,dx, if ∫01(f(x)−2g(x)) dx=6\int_{0}^{1}(f(x)-2g(x))\,dx=6, ∫01(2f(x)+2g(x)) dx=9\int_{0}^{1}(2f(x)+2g(x))\,dx=9
  1. Equations Addition: We have two equations:\newline11. ∫01(f(x)−2g(x))dx\int_{0}^{1}\left(f(x)-2g(x)\right)dx=66\newline22. ∫01(2f(x)+2g(x))dx\int_{0}^{1}\left(2f(x)+2g(x)\right)dx=99\newlineLet's add these two equations to eliminate g(x)g(x).
  2. Integrals Addition: Adding the integrals:\newline∫01((f(x)−2g(x)) dx)+∫01((2f(x)+2g(x)) dx)=6+9\int_{0}^{1}((f(x)-2g(x))\,dx) + \int_{0}^{1}((2f(x)+2g(x))\,dx) = 6 + 9\newline∫01(f(x)−2g(x)+2f(x)+2g(x)) dx=15\int_{0}^{1}(f(x)-2g(x) + 2f(x)+2g(x))\,dx = 15
  3. Integrand Simplification: Simplify the integrand: \int_{\(0\)}^{\(1\)}(f(x)\(-2g(x) + 22f(x)+22g(x))\,dx = \int_{00}^{11}(33f(x))\,dx
  4. Integral Calculation: Now we calculate the integral of 3f(x)3f(x) from 00 to 11: \newline∫01(3f(x)) dx=15\int_{0}^{1}(3f(x))\,dx = 15
  5. Integral Division: To find ∫01(f(x) dx)\int_{0}^{1}(f(x)\,dx), we divide both sides by 33:∫01(3f(x)) dx3=153\frac{\int_{0}^{1}(3f(x))\,dx}{3} = \frac{15}{3}∫01(f(x)) dx=5\int_{0}^{1}(f(x))\,dx = 5
  6. Integral Substitution: Now we need to find ∫01(g(x) dx)\int_{0}^{1}(g(x)\,dx). We can use the first given integral for that:\newline∫01((f(x)−2g(x)) dx)=6\int_{0}^{1}((f(x)-2g(x))\,dx) = 6\newlineWe already know ∫01(f(x) dx)=5\int_{0}^{1}(f(x)\,dx) = 5, so we substitute that in:\newline5−∫01(2g(x)) dx=65 - \int_{0}^{1}(2g(x))\,dx = 6
  7. Integral Solving: Solve for ∫01(2g(x)) dx\int_{0}^{1}(2g(x))\,dx:∫01(2g(x)) dx=5−6\int_{0}^{1}(2g(x))\,dx = 5 - 6∫01(2g(x)) dx=−1\int_{0}^{1}(2g(x))\,dx = -1
  8. Integral Division: Divide by 22 to find ∫01(g(x)) dx\int_{0}^{1}(g(x))\,dx: \newline∫01(2g(x)) dx2=−12\frac{\int_{0}^{1}(2g(x))\,dx}{2} = -\frac{1}{2}\newline∫01(g(x)) dx=−12\int_{0}^{1}(g(x))\,dx = -\frac{1}{2}
  9. Final Value Calculation: Finally, we find ∫01(f(x)−g(x)) dx\int_{0}^{1}(f(x)-g(x))\,dx using the values we found:\newline∫01f(x) dx−∫01g(x) dx=5−(−12)\int_{0}^{1}f(x)\,dx - \int_{0}^{1}g(x)\,dx = 5 - \left(-\frac{1}{2}\right)
  10. Final Value Calculation: Finally, we find ∫01(f(x)−g(x)) dx\int_{0}^{1}(f(x)-g(x))\,dx using the values we found:\newline∫01(f(x)) dx−∫01(g(x)) dx=5−(−12)\int_{0}^{1}(f(x))\,dx - \int_{0}^{1}(g(x))\,dx = 5 - \left(-\frac{1}{2}\right)Calculate the final value:\newline5−(−12)=5+125 - \left(-\frac{1}{2}\right) = 5 + \frac{1}{2}\newline5+12=5.55 + \frac{1}{2} = 5.5

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