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F(x)=5^(x) f(x)=F^(')(x) int_(0)^(3)f(x)dx=

F(x)=5x F(x)=5^{x} \newlinef(x)=F′(x) f(x)=F^{\prime}(x) \newline∫03f(x)dx= \int_{0}^{3} f(x) d x=

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Evaluate definite integrals using the power ruleright chevron icon

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Q. F(x)=5x F(x)=5^{x} \newlinef(x)=F′(x) f(x)=F^{\prime}(x) \newline∫03f(x)dx= \int_{0}^{3} f(x) d x=
  1. Integrate from 00 to 33: Now, we integrate f(x)f(x) from 00 to 33.\newline∫03ln⁡(5)⋅5x dx\int_{0}^{3} \ln(5) \cdot 5^{x} \, dx\newlineLet's pull out the constant ln⁡(5)\ln(5).\newlineln⁡(5)⋅∫035x dx\ln(5) \cdot \int_{0}^{3} 5^{x} \, dx
  2. Pull out constant ln⁡(5)\ln(5): To integrate 5x5^x, we use the formula ∫ax dx=axln⁡(a)+C\int a^x \, dx = \frac{a^x}{\ln(a)} + C.∫5x dx=5xln⁡(5)+C\int 5^x \, dx = \frac{5^x}{\ln(5)} + C Now we apply the limits from 00 to 33.ln⁡(5)×[53ln⁡(5)−50ln⁡(5)]\ln(5) \times \left[\frac{5^3}{\ln(5)} - \frac{5^0}{\ln(5)}\right]
  3. Apply integration formula: Simplify the expression.\newlineln⁡(5)×[125ln⁡(5)−1ln⁡(5)]\ln(5) \times \left[\frac{125}{\ln(5)} - \frac{1}{\ln(5)}\right]\newlineThis simplifies to 125−1=124125 - 1 = 124.

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