Question

How do you evaluate $\dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} $?

Answer 1

Hint: We can use the fundamental theorem of calculus to evaluate a given term. First, use a property of definite integrals that can split the limits of the integral. Next, use the chain rule to get $x$ as the upper limit. The final step is to get everything back in terms of $x$.

Formula used: Fundamental Theorem of Calculus:
If $f\left( x \right)$ is continuous on $\left[ {a,b} \right]$ then,
$g\left( x \right) = \int_a^x {f\left( t \right)dt} $
is continuous on $\left[ {a,b} \right]$ and it is differentiable on $\left( {a,b} \right)$ and that,
$g'\left( x \right) = f\left( x \right)$
An alternate notation for the derivative portion of this is,
$\dfrac{d}{{dx}}\int_a^x {f\left( t \right)dt} = f\left( x \right)$

Complete step-by-step solution:
We have to find $\dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} $.
This one needs a little work before we can use the Fundamental Theorem of Calculus. The first thing to notice is that the Fundamental Theorem of Calculus requires the lower limit to be a constant and the upper limit to be the variable. So, using a property of definite integrals we can split the limits of the integral we just need to remember to add in a minus sign after we do that. Doing this gives,
$\dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} = \dfrac{d}{{dx}}\int_0^{{x^4}} {\sqrt {{t^2} + t} dt} - \dfrac{d}{{dx}}\int_0^x {\sqrt {{t^2} + t} dt} $
The next thing to notice is that the Fundamental Theorem of Calculus also requires an $x$ in the upper limit of integration and we’ve got ${x^4}$. To do this derivative we’re going to need the following version of the chain rule.
$\dfrac{d}{{dx}}\left( {g\left( u \right)} \right) = \dfrac{d}{{du}}\left( {g\left( u \right)} \right)\dfrac{{du}}{{dx}}$, where $u = f\left( x \right)$
So, if we let $u = {x^4}$ we use the chain rule to get,
$\dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} = \dfrac{d}{{du}}\int_0^u {\sqrt {{t^2} + t} dt} \dfrac{{du}}{{dx}} - \dfrac{d}{{dx}}\int_0^x {\sqrt {{t^2} + t} dt} $
$ \Rightarrow \dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} = \sqrt {{u^2} + u} \left( {4{x^3}} \right) - \sqrt {{x^2} + x} $
$ \Rightarrow \dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} = 4{x^3}\sqrt {{u^2} + u} - \sqrt {{x^2} + x} $
The final step is to get everything back in terms of $x$.
 $\dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} = 4{x^3}\sqrt {{{\left( {{x^4}} \right)}^2} + {x^4}} - \sqrt {{x^2} + x} $
$ \Rightarrow \dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} = 4{x^3} \times {x^2}\sqrt {{x^4} + 1} - \sqrt {{x^2} + x} $
$ \Rightarrow \dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} = 4{x^5}\sqrt {{x^4} + 1} - \sqrt {{x^2} + x} $

Hence, $\dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} = 4{x^5}\sqrt {{x^4} + 1} - \sqrt {{x^2} + x} $.

Note: We can also solve the given integral using property
$\dfrac{d}{{dx}}\int_{v\left( x \right)}^{u\left( x \right)} {f\left( t \right)dt} = \dfrac{d}{{dx}}\left( {\int_{v\left( x \right)}^a {f\left( t \right)dt} + \int_a^{u\left( x \right)} {f\left( t \right)dt} } \right)$
Or $\dfrac{d}{{dx}}\int_{v\left( x \right)}^{u\left( x \right)} {f\left( t \right)dt} = - v'\left( x \right)f\left( {v\left( x \right)} \right) + u'\left( x \right)f\left( {u\left( x \right)} \right)$
Here, $f\left( t \right) = \sqrt {{t^2} + t} $, $u\left( x \right) = {x^4}$ and $v\left( x \right) = x$.
Use property $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}},n \ne - 1$ to find the differentiation of $u$ and $v$.
$u'\left( x \right) = 4{x^3}$ and $v'\left( x \right) = 1$
Now, determine $f\left( {u\left( x \right)} \right)$ and $f\left( {v\left( x \right)} \right)$.
$f\left( {u\left( x \right)} \right) = f\left( {{x^4}} \right)$
$ \Rightarrow f\left( {u\left( x \right)} \right) = \sqrt {{{\left( {{x^4}} \right)}^2} + {x^4}} $
$ \Rightarrow f\left( {u\left( x \right)} \right) = {x^2}\sqrt {{x^4} + 1} $
Now, $f\left( {v\left( x \right)} \right) = f\left( x \right)$
$ \Rightarrow f\left( {v\left( x \right)} \right) = \sqrt {{x^2} + x} $
Now, putting all these values in $\dfrac{d}{{dx}}\int_{v\left( x \right)}^{u\left( x \right)} {f\left( t \right)dt} = - v'\left( x \right)f\left( {v\left( x \right)} \right) + u'\left( x \right)f\left( {u\left( x \right)} \right)$.
$\dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} = - \sqrt {{x^2} + x} + 4{x^3} \times {x^2}\sqrt {{x^4} + 1} $
$ \Rightarrow \dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} = 4{x^5}\sqrt {{x^4} + 1} - \sqrt {{x^2} + x} $
Hence, $\dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} = 4{x^5}\sqrt {{x^4} + 1} - \sqrt {{x^2} + x} $.