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Newton-Leibnitz's Formula - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
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Newton-Leibniz's Formula is considered one of the most asked concept.
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48 Questions around this concept.
Solve by difficulty
If $m$ and $n$ respectively are the number of local maximum and local minimum points of the function $f(x)=\int_0^{x^2} \frac{t^2-5 t+4}{2+e^t} d t$, then the ordered pair $(m, n)$ is
Let $f(x)$ be a non-negative continuous function such that the area bounded by the curve $y=f(x), x-a x i s$ and the ordinates $x=\frac{\pi}{4}$ and $x=\beta>\frac{\pi}{4}$ is $\left(\beta \sin \beta+\frac{\pi}{4} \cos \beta+\sqrt{2} \beta\right)$.Then $f\left(\frac{\pi}{2}\right)$ is
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If , then
equals:
If the function is differentiable then for
is equal to
$\lim _{x \rightarrow \frac{x}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$ is equal to :
A function f, continuous on the positive real axis has the property that $\int_1^{\mathrm{xy}} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\mathrm{y} \int_1^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}+\mathrm{x} \int_1^{\mathrm{y}} \mathrm{f}(\mathrm{t}) \mathrm{dt}, \forall \mathrm{x}>0, \mathrm{y}>0$. If $\mathrm{f}(1)=3$, then $\mathrm{f}^{\prime}(3)$ equals
$\lim_{x=0} \frac{\int_{0}^{x^{2}}{(\tan^{-1}t)^{2} dt }}{\int_{0}^{x^{4}}{(\sin\sqrt{t} dt})}$ is equal to
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Newton-Leibniz's Formula
If the functions u(x) and v(x) are defined and f(t) is a continuous function, then
$\frac{d}{d x}\left[\int_{\mathbf{u}(\mathbf{x})}^{\mathbf{v}(\mathbf{x})} \mathbf{f}(\mathbf{t}) \mathrm{dt}\right]=\mathbf{f}(\mathbf{v}(\mathbf{x})) \cdot \frac{\mathrm{d}}{\mathrm{dx}}\{\mathbf{v}(\mathbf{x})\}-\mathbf{f}(\mathbf{u}(\mathbf{x})) \cdot \frac{d}{d x}\{\mathbf{u}(\mathbf{x})\}$
Proof:
$\begin{array}{ll}\text { Let } & \frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{F}(\mathrm{x})\}=\mathrm{f}(\mathrm{x}) \\ \Rightarrow & \int_{\mathrm{u}(\mathrm{x})}^{\mathrm{v}(\mathrm{x})} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\mathrm{F}(\mathrm{v}(\mathrm{x}))-\mathrm{F}(\mathrm{u}(\mathrm{x})) \\ \Rightarrow & \frac{\mathrm{d}}{\mathrm{dx}}\left[\int_{\mathrm{u}(\mathrm{x})}^{\mathrm{v}(\mathrm{x})} \mathrm{f}(\mathrm{t}) \mathrm{dt}\right]=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{F}(\mathrm{v}(\mathrm{x}))-\mathrm{F}(\mathrm{u}(\mathrm{x}))) \\ \Rightarrow & \frac{\mathrm{d}}{\mathrm{dx}}\left[\int_{\mathrm{u}(\mathrm{x})}^{\mathrm{v}(\mathrm{x})} \mathrm{f}(\mathrm{t}) \mathrm{dt}\right]=\mathrm{F}^{\prime}(\mathrm{v}(\mathrm{x})) \frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{v}(\mathrm{x})\}-\mathrm{F}^{\prime}(\mathrm{u}(\mathrm{x})) \frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{u}(\mathrm{x})\} \\ \Rightarrow & \frac{\mathrm{d}}{\mathrm{dx}}\left[\int_{\mathrm{u}(\mathrm{x})}^{\mathrm{v}(\mathrm{x})} \mathrm{f}(\mathrm{t}) \mathrm{dt}\right]=\mathrm{f}(\mathrm{v}(\mathrm{x})) \frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{v}(\mathrm{x})\}-\mathrm{f}(\mathrm{u}(\mathrm{x})) \frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{u}(\mathrm{x})\}\end{array}$
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