How do I evaluate $\frac{d}{d x} \int_{5}^{x^{4}} \sqrt{t^{2} + t} d t$ $d/dx \int_5^(x^4) \sqrt{t^2 + t} dt$ ?

Question

How do I evaluate $\frac{d}{d x} \int_{5}^{x^{4}} \sqrt{t^{2} + t} d t$ $d/dx \int_5^(x^4) \sqrt{t^2 + t} dt$ ?

2 Answers

Impact of this question

6497 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-06T20:39:57

To solve $\frac{d}{d x} \int_{5}^{x^{4}} \sqrt{t^{2} + t} d t$ $d /dx \int_5^{x^4} sqrt{t^2+t} dt$ , we will use the Fundamental Theorem of Calculus.

$\frac{d}{d x} \int_{5}^{x^{4}} \sqrt{t^{2} + t} d t$ $d /dx \int_5^{x^4} sqrt{t^2+t} dt$
$= [\sqrt{{(x^{4})}^{2} + x^{4}}] \cdot (4 x^{3})$ $= [\sqrt{(x^4)^2+x^4}] * (4x^3)$ **don't forget to chain!!

Answer 2 · 2017-03-06T20:53:25

$= 4x^5 \ sqrt(x^4 + 1)$

Explanation:

The FTC tells us that, for constant $a$ and parameter $u$ :

$d/(du) int_a^u f(t) dt = f(u)$

Now for the chaining bit.

If in fact $u = u(x)$ , we can say that:

$d/(dcolor(red)(x)) int_a^(u(x)) f(t) dt$

$= d/(du) (int_a^(u) f(t) dt ) cdot (du)/(dx)$

$= f(u) cdot (du)/dx$

So for the specific question, where $u(x) = x^4$ :

$d/(dx) int_5^(x^4) sqrt(t^2 + t) \ dt$

$= sqrt((x^4)^2 + x^4) cdot d/dx (x^4)$

$= x^2\ sqrt(x^4 + 1) cdot 4x^3$

$= 4x^5 \ sqrt(x^4 + 1)$

Science

Math

Humanities

... and beyond

How do I evaluate $\frac{d}{d x} \int_{5}^{x^{4}} \sqrt{t^{2} + t} d t$ $d/dx \int_5^(x^4) \sqrt{t^2 + t} dt$ ?

2 Answers

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do I evaluate d/dx \int_5^(x^4) \sqrt{t^2 + t} dtddx∫x45√t2+tdtd/dx \int_5^(x^4) \sqrt{t^2 + t} dt?

2 Answers

Explanation:

Related questions

Impact of this question

How do I evaluate $\frac{d}{d x} \int_{5}^{x^{4}} \sqrt{t^{2} + t} d t$ $d/dx \int_5^(x^4) \sqrt{t^2 + t} dt$ ?