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Calculus / English

Ex 7: Second Fundamental Theorem of Calculus with Chain Rule

Mathispower4u · YouTube · 4:16

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Combine the fundamental theorem of calculus with the chain rule when an integral has a square-root upper limit.

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Chapters

0:00Differentiate an accumulated integral0:50Differentiate the endpoint too3:45Simplify after substitution

Learning script

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When the integrand is continuous on the relevant interval, differentiating an integral with a variable upper endpoint evaluates the integrand at that endpoint.

The endpoint is u=√t, so for t>0t>0 its derivative is 1/(2√t). Multiply this chain-rule factor by the integrand evaluated at u.

For the lesson’s integrand sin(x⁵)/x⁴ and constant lower endpoint 2, the derivative at t>0t>0 is sin⁡(t(5/2))/(2t(5/2))\sin (t^(5/2))/(2t^(5/2)). The reciprocal endpoint-derivative factor is essential.

Knowledge cards

01

Differentiate an accumulated integral

When the integrand is continuous on the relevant interval, differentiating an integral with a variable upper endpoint evaluates the integrand at that endpoint.

02

Differentiate the endpoint too

The endpoint is u=√t, so for t>0t>0 its derivative is 1/(2√t). Multiply this chain-rule factor by the integrand evaluated at u.

03

Simplify after substitution

For the lesson’s integrand sin(x⁵)/x⁴ and constant lower endpoint 2, the derivative at t>0t>0 is sin⁡(t(5/2))/(2t(5/2))\sin (t^(5/2))/(2t^(5/2)). The reciprocal endpoint-derivative factor is essential.

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  • Chain rule ApplicationAt 0:00
    Why this connection?

    When the integrand is continuous on the relevant interval, differentiating an integral with a variable upper endpoint evaluates the integrand at that endpoint. The endpoint is u=√t, so for t>0t>0 its derivative is 1/(2√t). Multiply this chain-rule factor by the integrand evaluated at u. For the lesson’s integrand sin(x⁵)/x⁴ and constant lower endpoint 2, the derivative at t>0t>0 is sin⁡(t(5/2))/(2t(5/2))\sin (t^(5/2))/(2t^(5/2)). The reciprocal endpoint-derivative factor is essential.