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g(x)=int_(0)^(x)sqrt(5+4cos t)dt g^(')(pi)=

g(x)=∫0x5+4cos⁡tdt g(x)=\int_{0}^{x} \sqrt{5+4 \cos t} d t \newlineg′(π)= g^{\prime}(\pi)=

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Evaluate definite integrals using the power ruleright chevron icon

Full solution

Q. g(x)=∫0x5+4cos⁡tdt g(x)=\int_{0}^{x} \sqrt{5+4 \cos t} d t \newlineg′(π)= g^{\prime}(\pi)=
  1. Apply Fundamental Theorem: We need to use the Fundamental Theorem of Calculus to find g′(x)g'(x). Since g(x)g(x) is an integral from 00 to xx, g′(x)g'(x) is just the integrand evaluated at xx.
  2. Find g′(x)g'(x): So, g′(x)=5+4cos⁡(x)g'(x) = \sqrt{5 + 4\cos(x)}.
  3. Evaluate at x=πx = \pi: Now we evaluate g′(x)g'(x) at x=πx = \pi: g′(π)=5+4cos⁡(π)g'(\pi) = \sqrt{5 + 4\cos(\pi)}.
  4. Substitute cos⁡(π)\cos(\pi): We know that cos⁡(π)=−1\cos(\pi) = -1, so plug that in: g′(π)=5+4(−1)g'(\pi) = \sqrt{5 + 4(-1)}.
  5. Simplify expression: Simplify the expression: g′(π)=5−4.g'(\pi) = \sqrt{5 - 4}.
  6. Calculate value: Calculate the value: g′(π)=1.g'(\pi) = \sqrt{1}.
  7. Final result: The square root of 11 is 11, so g′(π)=1g'(\pi) = 1.

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