VIT - VITEEE 2025
ApplyNational level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
Newton-Leibniz's Formula is considered one of the most asked concept.
57 Questions around this concept.
A continuous function $f: \mathbb{R} \rightarrow \mathbb{R}_{\text {satisfies the equation }} f(x)=x+\int_0^x f(t) d t$ Which of the following options is true?
Let for some function $\mathrm{y}=\mathrm{f}(\mathrm{x}), \int_0^x \mathrm{t} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\mathrm{x}^2 \mathrm{f}(\mathrm{x})$, $x>0$ and $f(2)=3$. Then $f(6)$ is equal to :
Let $f$ be a real valued continuous function defined on the positive real axis such that $\mathrm{g}(\mathrm{x})=\int_0^{\mathrm{x}} \mathrm{t} f(\mathrm{t}) \mathrm{dt}$. If $g\left(x^3\right)=x^6+x^7$, then value of $\sum_{r=1}^{15} f\left(r^3\right)$ is:
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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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