If f(x)= ∫(21,x) t^5 dt then f'(x).
my answer was 1/6 (21^6-x^6) it said it's wrong.
You must be Mily from below.
same thing as before
∫ t^5 dt from t=21 to t=x
= (1/6)t^6 from t-21 to t=x
= (1/6)x^6 - (1/6)(21)^6
= (1/6)(x^6 - 21^6)
what about f ' (x)?
i did with (1/6)(x^6-21^6) but its wrong
f(x)=∫(21,x) t^5
f'(x)= x^5
i font know
but thanks
i got it, it correct answer is f'(x)=-x^5.
To find the derivative of the function f(x) = ∫(21,x) t^5 dt, we can use the Fundamental Theorem of Calculus. According to this theorem, if a function F(x) is defined as the integral of another function f(t), then the derivative of F(x) with respect to x is equal to f(x).
In this case, we have f(x) = ∫(21,x) t^5 dt. To find f'(x), we differentiate the integral with respect to x while treating the upper limit as a constant:
f'(x) = d/dx ∫(21,x) t^5 dt
To differentiate this integral, we can use the Second Fundamental Theorem of Calculus, which states that if the function f(t) is continuous on the interval [a, b] and F(x) is an antiderivative of f(x) on [a,b], then:
d/dx ∫(a,x) f(t) dt = f(x)
Now, let's find an antiderivative of t^5. The antiderivative of t^n is (1/(n+1)) * t^(n+1), so:
∫ t^5 dt = (1/(5+1)) * t^(5+1) = (1/6) * t^6
Therefore, an antiderivative of t^5 is (1/6) * t^6.
Using the Second Fundamental Theorem of Calculus, we replace t^5 with (1/6) * t^6:
f'(x) = d/dx ∫(21,x) (1/6) * t^6 dt = (1/6) * x^6
Hence, the correct answer for f'(x) is (1/6) * x^6.