Can you apply the fundamental theorem of calculus with the variable inside the integrand?

For example: $\frac{d}{dx}\left({\int}_{a}^{x}xf(t)dt\right)$

For example: $\frac{d}{dx}\left({\int}_{a}^{x}xf(t)dt\right)$

Leland Morrow
2022-07-01
Answered

Can you apply the fundamental theorem of calculus with the variable inside the integrand?

For example: $\frac{d}{dx}\left({\int}_{a}^{x}xf(t)dt\right)$

For example: $\frac{d}{dx}\left({\int}_{a}^{x}xf(t)dt\right)$

You can still ask an expert for help

hopeloothab9m

Answered 2022-07-02
Author has **25** answers

By definition

$\frac{d}{dx}{\int}_{a}^{x}f(x,t)dt=\underset{\mathrm{\Delta}x\to 0}{lim}\frac{1}{\mathrm{\Delta}x}[{\int}_{a}^{x+\mathrm{\Delta}x}f(x+\mathrm{\Delta}x,t)dt-{\int}_{a}^{x}f(x,t)dt]$

$=\underset{\mathrm{\Delta}x\to 0}{lim}\frac{1}{\mathrm{\Delta}x}[{\int}_{a}^{x+\mathrm{\Delta}x}(f(x,t)+\frac{\mathrm{\partial}f}{\mathrm{\partial}x}\mathrm{\Delta}x)dt-{\int}_{a}^{x}f(x,t)dt]$

$=\underset{\mathrm{\Delta}x\to 0}{lim}\frac{1}{\mathrm{\Delta}x}[{\int}_{a}^{x+\mathrm{\Delta}x}f(x,t)dt-{\int}_{a}^{x}f(x,t)dt+{\int}_{a}^{x+\mathrm{\Delta}x}\frac{\mathrm{\partial}f}{\mathrm{\partial}x}\mathrm{\Delta}xdt]$

$=\underset{\mathrm{\Delta}x\to 0}{lim}\frac{1}{\mathrm{\Delta}x}[{\int}_{x}^{x+\mathrm{\Delta}x}f(x,t)dt+{\int}_{a}^{x+\mathrm{\Delta}x}\frac{\mathrm{\partial}f}{\mathrm{\partial}x}\mathrm{\Delta}xdt]$

$=f(x,x)+\underset{\mathrm{\Delta}x\to 0}{lim}{\int}_{a}^{x+\mathrm{\Delta}x}\frac{\mathrm{\partial}f}{\mathrm{\partial}x}dt$

$=f(x,x)+{\int}_{a}^{x}\frac{\mathrm{\partial}}{\mathrm{\partial}x}f(x,t)dt$

In conclusion, you will only get the first term if you apply the FTOC blindly. Because the integrand contains $x$, you will have the second extra term.

$\frac{d}{dx}{\int}_{a}^{x}f(x,t)dt=\underset{\mathrm{\Delta}x\to 0}{lim}\frac{1}{\mathrm{\Delta}x}[{\int}_{a}^{x+\mathrm{\Delta}x}f(x+\mathrm{\Delta}x,t)dt-{\int}_{a}^{x}f(x,t)dt]$

$=\underset{\mathrm{\Delta}x\to 0}{lim}\frac{1}{\mathrm{\Delta}x}[{\int}_{a}^{x+\mathrm{\Delta}x}(f(x,t)+\frac{\mathrm{\partial}f}{\mathrm{\partial}x}\mathrm{\Delta}x)dt-{\int}_{a}^{x}f(x,t)dt]$

$=\underset{\mathrm{\Delta}x\to 0}{lim}\frac{1}{\mathrm{\Delta}x}[{\int}_{a}^{x+\mathrm{\Delta}x}f(x,t)dt-{\int}_{a}^{x}f(x,t)dt+{\int}_{a}^{x+\mathrm{\Delta}x}\frac{\mathrm{\partial}f}{\mathrm{\partial}x}\mathrm{\Delta}xdt]$

$=\underset{\mathrm{\Delta}x\to 0}{lim}\frac{1}{\mathrm{\Delta}x}[{\int}_{x}^{x+\mathrm{\Delta}x}f(x,t)dt+{\int}_{a}^{x+\mathrm{\Delta}x}\frac{\mathrm{\partial}f}{\mathrm{\partial}x}\mathrm{\Delta}xdt]$

$=f(x,x)+\underset{\mathrm{\Delta}x\to 0}{lim}{\int}_{a}^{x+\mathrm{\Delta}x}\frac{\mathrm{\partial}f}{\mathrm{\partial}x}dt$

$=f(x,x)+{\int}_{a}^{x}\frac{\mathrm{\partial}}{\mathrm{\partial}x}f(x,t)dt$

In conclusion, you will only get the first term if you apply the FTOC blindly. Because the integrand contains $x$, you will have the second extra term.

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Suppose $F(x)={\int}_{3x+8}^{{x}^{2}+5x+1}{\mathrm{csc}}^{2}\left(t\right)dt$. How would one find ${F}^{\prime}(x)$ using the first fundamental theorem of calculus?

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Finding $f(x)$ in ${\mathrm{cos}}^{2}(x)f(x)={x}^{2}-2{\int}_{1}^{x}\mathrm{sin}(t)\mathrm{cos}(t)f(t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t$

asked 2022-06-20

If $f(x)$ is even, then what can we say about:

${\int}_{-2}^{2}f(x)dx$

If $f(x)$ is odd, then what can we say about

${\int}_{-2}^{2}f(x)dx$

Are they both zero? For the first one if its even wouldn't this be the same as

${\int}_{a}^{a}f(x)dx=0$

Now if its odd $f(-x)=-f(x)$. Would FTOC make this zero as well?

${\int}_{-2}^{2}f(x)dx$

If $f(x)$ is odd, then what can we say about

${\int}_{-2}^{2}f(x)dx$

Are they both zero? For the first one if its even wouldn't this be the same as

${\int}_{a}^{a}f(x)dx=0$

Now if its odd $f(-x)=-f(x)$. Would FTOC make this zero as well?

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Can use FToC to evaluate $\underset{x\to \mathrm{\infty}}{lim}\frac{{\int}_{0}^{x}\phantom{\rule{mediummathspace}{0ex}}f\left(t\right)dt}{{x}^{2}}$?

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Let $f:[a,b]\to \mathbb{R}$ be monotone increasing, $F:[a,b]\to \mathbb{R}$ such that $F(x):={\int}_{[a,x]}f$ and ${x}_{0}\in [a,b]$. Then $F$ is differentiable at ${x}_{0}$ if and only if $f$ is continuous at ${x}_{0}$

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First fundamental theorem of calculus uses a function

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$F(x)$ is an antiderivative of function $f$.

So what closed interval is considered when we take indefinite integral?

$F(x)=\text{}{\int}_{a}^{x}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt$ for $f$ a continuous function between $[a,b]$ where $x$ is between $[a,b]$

$F(x)$ is an antiderivative of function $f$.

So what closed interval is considered when we take indefinite integral?

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Derivative of a function and an integral

$\frac{d}{dx}({x}^{6}({\int}_{0}^{sinx}\sqrt{t}dt))$

$\frac{d}{dx}({x}^{6}({\int}_{0}^{sinx}\sqrt{t}dt))$