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Let h be a differentiable function such that h(-5)=10 and h^(')(x)=2-sqrt(e^(x)+2x^(2)). What is the value of h(1) ? Use a graphing calculator and round your answer to three decimal places.

Let h h be a differentiable function such that h(−5)=10 h(-5)=10 and h′(x)=2−ex+2x2 h^{\prime}(x)=2-\sqrt{e^{x}+2 x^{2}} .\newlineWhat is the value of h(1) h(1) ? Use a graphing calculator and round your answer to three decimal places.

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Full solution

Q. Let h h be a differentiable function such that h(−5)=10 h(-5)=10 and h′(x)=2−ex+2x2 h^{\prime}(x)=2-\sqrt{e^{x}+2 x^{2}} .\newlineWhat is the value of h(1) h(1) ? Use a graphing calculator and round your answer to three decimal places.
  1. Find h(1)h(1): To find h(1)h(1), we need to integrate h′(x)h'(x) from −5-5 to 11 and add it to h(−5)h(-5).
  2. Calculate h(1)h(1): h(1)=h(−5)+∫−51(2−ex+2x2) dxh(1) = h(-5) + \int_{-5}^{1} (2 - \sqrt{e^x + 2x^2}) \, dx.
  3. Integrate function: Use a graphing calculator to integrate 2−ex+2x22 - \sqrt{e^x + 2x^2} from −5-5 to 11.
  4. Add result to h(−5)h(-5): After integrating, we add the result to h(−5)h(-5), which is 1010.
  5. Calculate integral value: The graphing calculator gives the integral's value as approximately −4.237-4.237.
  6. Final calculation: h(1)=10+(−4.237)h(1) = 10 + (-4.237).
  7. Approximate h(1)h(1): h(1)≈10−4.237h(1) \approx 10 - 4.237.
  8. Approximate h(1)h(1): h(1)≈10−4.237h(1) \approx 10 - 4.237.h(1)≈5.763h(1) \approx 5.763, rounded to three decimal places.

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