Chapter 4: Integration

Section 4.3: Fundamental Theorem of Calculus and the Indefinite Integral

Example 4.3.3

$$ Obtain an explicit rule for the function $g\left(x\right)\={\int }_0^x\frac{1}{\sqrt{1\+x^2}} \mathit{\ⅆ}x$; then show $g\prime \left(x\right)$ is the integrand evaluated at $x$. $$ Solution $$ •  Control-drag the equation $g\left(x\right)\=\dots$  •  Context Panel: Assign Function $g\left(x\right)\={\int }_0^x\frac{1}{\sqrt{1\+t^2}} \mathit{\ⅆ}t$$\stackrel{\text{assign as function}}{\to }$$g$ •  Write the derivative notation $g\prime \left(x\right)$ Context Panel: Evaluate and Display Inline $g\prime \left(x\right)$ = $\frac{1}{\sqrt{x^2\+1}}$$$ The explicit rule for $g\left(x\right)$ and its derivative •  Write $g\left(x\right)$ and press the Enter key. (This displays the explicit rule for $g\left(x\right)$.)   •  Context Panel: Differentiate≻With Respect To≻$x$ $g\left(x\right)$ $\mathrm{arcsinh}\left(x\right)$ $\stackrel{\text{differentiate w.r.t. x}}{\to }$ $\frac{1}{\sqrt{x^2\+1}}$ $$ As per the Fundamental Theorem of Calculus, the definite integral defines the antiderivative, so the derivative is the integrand.   The antiderivative of the integrand $1\/\sqrt{1\+t^2}$ is found by consulting Table 3.10.1 , or by relying on Maple.

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