The equation of the tangent to the $$curvey=$$∫$$xx2$$ log⁡tdt at $$x=2$$ is

Correct option is 4y+8 log 2+2=(7log 2)x

Questions

The equation of the tangent to the curve
$y = \int_{x}^{x^{2}} \log t d t at x = 2$ is

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y–6 log 2=7log2(x–2)

y−log⁡2e1/3=(log⁡2)x

y−8log⁡2e−1/3=5log⁡2x

y+8 log 2+2=(7log 2)x

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detailed solution

Correct option is D

y=∫xx2 log⁡tdt=(tlog⁡t−t)xx2=2x2−xlog⁡x−x2+x⇒y(2)=6log⁡2−2Also, y′(2)=7log⁡2Thus, equation of tangent at x = 2 isy−(6log⁡2−2)=(7log⁡2)(x−2) or y+8log⁡2+2=(7log⁡2)(x)

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The equation of the tangent to the curvey=∫xx2 log⁡tdt at x=2 is

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The equation of the tangent to the curve
$y = \int_{x}^{x^{2}} \log t d t at x = 2$ is