Integration (Area under a graph)

Let $A(r)=$ Area bounded by lines $f(x), x-axis, x=0$ and $x=r \Leftrightarrow$ Area between graph f(x) and x-axis from $x=0$ to $x=r$

$A$ can be approximated as a trapezium if $\delta x$ is very small

$\delta A \approx \frac{f(x + \delta x) + f(x)}{2} \cdot \ \delta x$

$\frac{\delta A}{\delta x} \approx \frac{f(x + \delta x) + f(x)}{2}$

$\frac{dA}{dx} = \lim_{\delta x \rightarrow 0} {\frac{\delta A}{\delta x}} = \frac{f(x+0) +f(x)}{2}$

$\frac{dA}{dx} = \frac{2f(x)}{2}$

$\frac{dA}{dx} = f(x)$

$A = \int_{0} ^r f(x) dx$

#Calculus

Easy Math Editor

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Integration (Area under a graph)

δA≈f(x+δx)+f(x)2⋅ δx \delta A \approx \frac{f(x + \delta x) + f(x)}{2} \cdot \ \delta x δA≈2f(x+δx)+f(x)​⋅ δx

δAδx≈f(x+δx)+f(x)2 \frac{\delta A}{\delta x} \approx \frac{f(x + \delta x) + f(x)}{2} δxδA​≈2f(x+δx)+f(x)​

dAdx=lim⁡δx→0δAδx=f(x+0)+f(x)2 \frac{dA}{dx} = \lim_{\delta x \rightarrow 0} {\frac{\delta A}{\delta x}} = \frac{f(x+0) +f(x)}{2} dxdA​=limδx→0​δxδA​=2f(x+0)+f(x)​

dAdx=2f(x)2 \frac{dA}{dx} = \frac{2f(x)}{2} dxdA​=22f(x)​

dAdx=f(x) \frac{dA}{dx} = f(x) dxdA​=f(x)

A=∫0rf(x)dx A = \int_{0} ^r f(x) dx A=∫0r​f(x)dx