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F(x)=sqrt(x+7) f(x)=F^(')(x) int_(2)^(9)f(x)dx=

F(x)=x+7 F(x)=\sqrt{x+7} \newlinef(x)=F′(x) f(x)=F^{\prime}(x) \newline∫29f(x)dx= \int_{2}^{9} f(x) d x=

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Evaluate definite integrals using the power ruleright chevron icon

Full solution

Q. F(x)=x+7 F(x)=\sqrt{x+7} \newlinef(x)=F′(x) f(x)=F^{\prime}(x) \newline∫29f(x)dx= \int_{2}^{9} f(x) d x=
  1. Identify Function: F(x)=x+7F(x) = \sqrt{x+7}, so let's find F′(x)F'(x) which is f(x)f(x).
  2. Find Derivative: Using the chain rule, F′(x)=12x+7⋅(x+7)′=12x+7⋅1F'(x) = \frac{1}{2\sqrt{x+7}} \cdot (x+7)' = \frac{1}{2\sqrt{x+7}} \cdot 1.
  3. Calculate Integral: Now we have f(x)=12x+7f(x) = \frac{1}{2\sqrt{x+7}}. Let's integrate f(x)f(x) from 22 to 99.
  4. Evaluate Upper Limit: ∫2912x+7 dx=[x+7]\int_{2}^{9} \frac{1}{2\sqrt{x+7}} \, dx = [\sqrt{x+7}] from 22 to 99.
  5. Evaluate Lower Limit: Plug in the upper limit: 9+7=16=4\sqrt{9+7} = \sqrt{16} = 4.
  6. Subtract Limits: Plug in the lower limit: 2+7=9=3\sqrt{2+7} = \sqrt{9} = 3.
  7. Subtract Limits: Plug in the lower limit: 2+7=9=3\sqrt{2+7} = \sqrt{9} = 3.Now subtract the lower limit from the upper limit: 4−3=14 - 3 = 1.

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