Computing integral using Fundamental Theorem of Calculus Part I

Question

Question: Find the derivative of $\int^x_3{\sin^3t}dt$

Since $f(x)$ is a trig polynomial, $f(x)$ is continuous, so I applied FTOC I.

$\frac{d}{dt}(\int^x_3{\sin^3t}dt$)

= $F'(x)$

= $\sin^3x$ (substituted x into t)

Apparently this is not the answer, can someone tell me what I'm doing wrong here?

What's the question? $\int_3^x \sin^2(t) , dt$ is not a question, it's just an expression. — littleO, Jun 09 '18 at 20:43
Sorry for being unclear, I added a statement to the question. — Michael Johnson, Jun 09 '18 at 20:45
how come when I integrate it using the integral-calculator.com it tells me the integral is cos^3t/3 - cos(t) | upper = x, lower = 3 (using FTOC II) — Michael Johnson, Jun 09 '18 at 20:47
@MichaelJohnson The integral calculator is actually taking the integral, not the derivative. — 高田航, Jun 09 '18 at 20:47
@MichaelJohnson Your question is to finding the derivative of the integral . — S.H.W, Jun 09 '18 at 20:48
For finding $\int^x_3{sin^3t}dt$ use $\sin 3x = 3\sin x -4\sin^3 x$ . — S.H.W, Jun 09 '18 at 20:50
My understanding is kind of weak in this area, but the derivative of an integral is just another way of solving an integral, no? So shouldn't the answers be equivalent in some way? — Michael Johnson, Jun 09 '18 at 20:51
See this : https://math.stackexchange.com/questions/37656/how-to-calculate-the-derivative-of-this-integral — S.H.W, Jun 09 '18 at 20:54

score 4 · Accepted Answer · answered Jun 09 '18 at 20:53

It seems that you are confused about what the problem is actually asking. According to your post, we want to evaluate $$\frac{d}{dx}\Big\lbrack\int_3^x\sin^3(t)dt\Big\rbrack$$ By the fundamental theorem of calculus, we can "replace" the $t$ and $x$, to arrive at the answer $$\sin^3(x)$$

However, if the question was to simply evaluate the integral $$\int_3^x\sin^3(t)dt$$ then we have $$\dfrac{\cos^3\left(x\right)-3\cos\left(x\right)}{3}-\dfrac{\cos^3\left(3\right)-3\cos\left(3\right)}{3}$$

The two procedures above are very different, since taking the derivative of an integral (i.e. applying $\frac{d}{dx}$) is to simply "undo" the integrating process and be left with the original function, whereas integrating the function is to find the antiderivative.

aH, thanks for clearing that up – Michael Johnson Jun 09 '18 at 20:56 — Michael Johnson, Jun 09 '18 at 20:56

score 0 · Answer 2 · answered Jun 09 '18 at 21:19

The Fundamental Theorem of Calculus states that if $$g(x) = \int_{a}^{f(x)} h(t)~{\rm d}t$$ where $a$ is any constant, then $$g'(x) = h(f(x)) \cdot f'(x)$$ Using this with the integral, $g(x) = g(x)$, $f(x) = x$, and $h(x) = \sin^3 t$. We may now plug in these values. $$g'(x) = \sin^3(x)$$ So, $$\frac{d}{dx}\Big\lbrack\int_3^x\sin^3(t)dt\Big\rbrack=\sin^3(x)$$

Computing integral using Fundamental Theorem of Calculus Part I

2 Answers2