Fundamental Theorem of Calculus Part I

Problem 183: Let

(1) $\begin{equation*} y = \int_{1+3x^2}^{4} \frac{1}{2+e^t} \ dt.\end{equation*}$

Find $dy/dx$ .

Solution: We will use the Fundamental Theorem of Calculus Part I to find the derivative of the integral. That is,

(2) $\begin{align*} y & = \int_{2+3x^2}^{4} \frac{1}{2+e^t} \ dt \\ \frac{dy}{dx} & = \frac{d}{dx} \left( \int_{2+3x^2}^{4} \frac{1}{2+e^t} \ dt \right) \\ & = \frac{d}{dx} \left( -\int_{4}^{2+3x^2} \frac{1}{2+e^t} \ dt \right) \qquad \qquad \left(\int_{a}^{b} = - \int_{b}^{a} \right) \\ & = - \frac{d}{dx} \left( \int_{4}^{2+3x^2} \frac{1}{1+e^t} \ dt \right) \\ & = - \frac{1}{2+e^{1+3x^2}} \cdot \frac{d}{dx}(2+3x^2) \\ & = - \frac{6x}{2+e^{(1+3x^2)}}. \end{align*}$

That is,

(3) $\begin{equation*} \frac{dy}{dx} = - \frac{6x}{2+e^{(1+3x^2)}} \end{equation*}$

One Problem at a Time

Fundamental Theorem of Calculus Part I

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