39Non-integrable function...it should be square root power.
Let
f(x)=(x2+4x)1/3
and let g(x) be an antiderivative of f(x). Then if
g(5) = 7
find
g(1)
-
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12 Answers
3939It's a cube root Ray. It can't be integrated. I don't think you've done integral calculus as yet...the "anti-derivative" is also known as the integral.
39Superb question hai srk...solve hi nahi kiya jaa sakta(unless you are Gallardo, Anant sir, or the like). Next time give us an integrable function, okay? [1]
1well pritish u need to reshuffle ur calculus.anyhow here's solution
Since g(x) is an antiderivative of f(x), we have
g '(x) = f(x)
or g(x)=∫(t2+4t)1/3dt,taking limits L-a and U-x
None of the regular techniques of integration will work on this integral. Even the computer cannot solve this explicitly. Instead of integrating, we let
h(x) = g(x) - 7
Then h(x) is also an antiderivative of f(x) and
h(5) = 0
We can write
h(x)=∫(t2+4t)1/3dt,take limits L-5,U-x
Notice that when we plug in 5 for x, we get 0 as required, since the upper and lower limits are equal. Now use a calculator to easily find
h(1)=)=∫(t2+4t)1/3dt=-10.88222 take limits l-5,u-1
finally since
h(x) = g(x) - 7
it follows that
g(x) = h(x) + 7
and that
g(1) = h(1) + 7 = -10.88222 + 7 = -3.88222
39"Now use a calculator to easily find.."
lol calculator kaun dega tujhe exam mein?
1@ srk bhai..wait toh karte..i was trying!![3][1]
Still, good one. But are we allowed a calculator?
39srk..do you understand your own solution? Bas itna bata do. [1]