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Question: How do you evaluate \(\dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} \)?...

How do you evaluate ddx∫xx4t2+tdt\dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} ?

Explanation

Solution

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 xx as the upper limit. The final step is to get everything back in terms of xx.

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

Complete step-by-step solution:
We have to find ddx∫xx4t2+tdt\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,
ddx∫xx4t2+tdt=ddx∫0x4t2+tdt−ddx∫0xt2+tdt\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 xx in the upper limit of integration and we’ve got x4{x^4}. To do this derivative we’re going to need the following version of the chain rule.
ddx(g(u))=ddu(g(u))dudx\dfrac{d}{{dx}}\left( {g\left( u \right)} \right) = \dfrac{d}{{du}}\left( {g\left( u \right)} \right)\dfrac{{du}}{{dx}}, where u=f(x)u = f\left( x \right)
So, if we let u=x4u = {x^4} we use the chain rule to get,
ddx∫xx4t2+tdt=ddu∫0ut2+tdtdudx−ddx∫0xt2+tdt\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}
⇒ddx∫xx4t2+tdt=u2+u(4x3)−x2+x\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}
⇒ddx∫xx4t2+tdt=4x3u2+u−x2+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 xx.
ddx∫xx4t2+tdt=4x3(x4)2+x4−x2+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}
⇒ddx∫xx4t2+tdt=4x3×x2x4+1−x2+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}
⇒ddx∫xx4t2+tdt=4x5x4+1−x2+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, ddx∫xx4t2+tdt=4x5x4+1−x2+x\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
ddx∫v(x)u(x)f(t)dt=ddx(∫v(x)af(t)dt+∫au(x)f(t)dt)\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 ddx∫v(x)u(x)f(t)dt=−v′(x)f(v(x))+u′(x)f(u(x))\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(t)=t2+tf\left( t \right) = \sqrt {{t^2} + t} , u(x)=x4u\left( x \right) = {x^4} and v(x)=xv\left( x \right) = x.
Use property ddx(xn)=nxn−1,n≠−1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}},n \ne - 1 to find the differentiation of uu and vv.
u′(x)=4x3u'\left( x \right) = 4{x^3} and v′(x)=1v'\left( x \right) = 1
Now, determine f(u(x))f\left( {u\left( x \right)} \right) and f(v(x))f\left( {v\left( x \right)} \right).
f(u(x))=f(x4)f\left( {u\left( x \right)} \right) = f\left( {{x^4}} \right)
⇒f(u(x))=(x4)2+x4\Rightarrow f\left( {u\left( x \right)} \right) = \sqrt {{{\left( {{x^4}} \right)}^2} + {x^4}}
⇒f(u(x))=x2x4+1\Rightarrow f\left( {u\left( x \right)} \right) = {x^2}\sqrt {{x^4} + 1}
Now, f(v(x))=f(x)f\left( {v\left( x \right)} \right) = f\left( x \right)
⇒f(v(x))=x2+x\Rightarrow f\left( {v\left( x \right)} \right) = \sqrt {{x^2} + x}
Now, putting all these values in ddx∫v(x)u(x)f(t)dt=−v′(x)f(v(x))+u′(x)f(u(x))\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).
ddx∫xx4t2+tdt=−x2+x+4x3×x2x4+1\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}
⇒ddx∫xx4t2+tdt=4x5x4+1−x2+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, ddx∫xx4t2+tdt=4x5x4+1−x2+x\dfrac{d}{{dx}}\int_x^{{x^4}} {\sqrt {{t^2} + t} dt} = 4{x^5}\sqrt {{x^4} + 1} - \sqrt {{x^2} + x} .