Finding the derivative of an integral: int[1,sqrt(x)] 18t^9 dt

khbeast21

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Evaluate the definite integral:

\(\displaystyle 13)\, \)\(\displaystyle \displaystyle \int_{-\pi/2}^{\pi/2}\, \)\(\displaystyle (\cos(x)\, +\, 5)\, dx\)

. . . . .\(\displaystyle \mbox{A) }\, 2\, +\, 5\pi\). . . . .B) 0. . . . .C) 7. . . . .\(\displaystyle \mbox{D) }\, 5\pi\)



Find dy/dx.

\(\displaystyle 14)\, \)\(\displaystyle \displaystyle \int_1^{\sqrt{\strut x\,}}\,\)\(\displaystyle 18t^9\, dt\)

. . . . .\(\displaystyle \mbox{A) }\, 12x^6\). . . . .\(\displaystyle \mbox{B) }\, \dfrac{9}{5}\, x^6\, -\, \dfrac{9}{5}\). . . . .\(\displaystyle \mbox{C) }\, 18x^{9/2}\). . . . .\(\displaystyle \mbox{D) }\, 9x^4\)



Evaluate the integral.

\(\displaystyle 15)\, \)\(\displaystyle \displaystyle \int_{1/5}^3\, \)\(\displaystyle \left(5\, -\, \dfrac{1}{x}\right)\, dx\)
 

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Did you read the RULES? If so where is your own effort?
 
I got to 9/4(t^10)+C, but I don't know what to do after that.

Okay, so you've taken the indefinite integral of the given function. I'd note, however, that's not quite the correct answer, though. The power rule says:

\(\displaystyle \int t^adt=\frac{t^{a+1}}{a+1},\:a\ne -1\)

Applying that, we have:

\(\displaystyle \int 18t^9dt=18\cdot \int \:t^9dt=??\)

Do you see now why having a denominator of 4 makes no sense? Now, that said, once you've got the correct answer, you'll need to account for the bounds of the integral that you're differentiating. Apply the fundamental theorem of calculus:

\(\displaystyle \int _a^b\:f\left(t\right)dt=F\left(b\right)-F\left(a\right)\) where F(t) is the indefinite integral ("anti-derivative") of f(t)

You know what F(t) is, so what happens if you apply the rule? That will give you a function of x, which you can then use the standard derivative rules to derivate.
 
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