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

g(x)=∫0x5+4tan⁡tdt g(x)=\int_{0}^{x} \sqrt{5+4 \tan t} d t \newlineg′(π4)= g^{\prime}\left(\frac{\pi}{4}\right)=

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Evaluate definite integrals using the power ruleright chevron icon

Full solution

Q. g(x)=∫0x5+4tan⁡tdt g(x)=\int_{0}^{x} \sqrt{5+4 \tan t} d t \newlineg′(π4)= g^{\prime}\left(\frac{\pi}{4}\right)=
  1. Apply Fundamental Theorem of Calculus: Use the Fundamental Theorem of Calculus which states that if g(x)=∫axf(t) dtg(x) = \int_{a}^{x} f(t) \, dt, then g′(x)=f(x)g'(x) = f(x).
  2. Evaluate integrand at x=π4x = \frac{\pi}{4}: Evaluate the integrand at x=π4x = \frac{\pi}{4}: f(π4)=5+4tan⁡(π4)f\left(\frac{\pi}{4}\right) = \sqrt{5 + 4\tan\left(\frac{\pi}{4}\right)}.
  3. Substitute into integrand: Since tan⁡(π4)=1\tan(\frac{\pi}{4}) = 1, substitute into the integrand: 5+4(1)=5+4\sqrt{5 + 4(1)} = \sqrt{5 + 4}.
  4. Simplify the expression: Simplify the expression: 9=3\sqrt{9} = 3.
  5. Final result: Therefore, g′(π4)=3g'(\frac{\pi}{4}) = 3.

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