Using Part 1 of the Fundamental Theorem of Calculus

Find \(\dfrac{d}{dx}\int_{4}^{3x^{2}+1}\sqrt{e^{t}+t}\,dt\).

Solution The upper limit of integration is a function of \(x\), so we use the Chain Rule along with Part 1 of the Fundamental Theorem of Calculus.

Let \(y=\int_{4}^{3x^{2}+1}\sqrt{e^{t}+t}\,dt\) and \(u(x) =3x^{2}+1.\) Then \(y=\int_{4}^{u}\sqrt{e^{t}+t}\,dt\) and \[ \begin{eqnarray*} \dfrac{d}{dx}\int_{4}^{3x^{2}+1}\sqrt{e^{t}+t}\,dt &=& \dfrac{dy}{dx} \underset{\underset{{\color{#0066A7}{\hbox{Chain Rule}}}}{\color{#0066A7}{\uparrow}}}{=} \dfrac{dy}{du}\cdot \dfrac{du}{dx}= \left[ \dfrac{d}{du}\int_{4}^{u}\sqrt{e^{t}+t}\,dt\right] \cdot \dfrac{du}{dx} \\ & \underset{\underset{\color{#0066A7}{{\hbox{Use the Fundamental Theorem}}}} {\color{#0066A7}{\uparrow}}}{=}& \sqrt{e^{u}+u}\cdot \dfrac{du}{dx} \underset{\underset{{\color{#0066A7}{{\hbox{\({u=3x}^{2}+1\); \(\dfrac{du}{dx}=6x\)}}}}{\hbox{\( \)}}}{\color{#0066A7}{\uparrow}}}{=}\sqrt{e^{( 3x^{2}+1) }+3x^{2}+1}\cdot 6x \\ \end{eqnarray*} \]