State the Fundamental Theorem of Calculus, and use it to differentiate,
$$F(x) = \int_1^{x^3} \cos(t^4)dt_.$$
$\endgroup$ 5Thoughts: The fundamental theorem of calculus states if a function $f$ is continuous on a closed interval $[a,b]$ and $F$ is the antiderivative of $f$ on the interval $[a,b]$
2 Answers
$\begingroup$The Fundamental Theorem of Calculus states: Let $f(x)$ is a continuous function on an open interval $I$, and $a \in I$. Then the following function is an antiderivative of $f(x)$ on $I$: $$A(x) = \int_a^x f(t)dt_.$$
Since $\cos(t^4)$ is continuous everywhere, we can apply FTC. Let $$F(x) = \int_1^{x^3} \cos(t^4) dt \\ A(x) = \int_1^x \cos(t^4)dt_.$$
By FTC, $A'(x) = \cos(x^4)$. Now notice that $F(x) = A(x^3)$. Therefore:
\begin{align*} F'(x) &= A'(x^3)(3x^2) \\ &= \cos((x^3)^4)(3x^2) \\ &= \cos(x^{12})(3x^2). \end{align*}
$\endgroup$ 3 $\begingroup$Let$f(t)=\cos(t^4), g(x)=x^3, $
$F(X)=\int_0^X f(t)dt$ and $G(x)=F(g(x))$.
$f$ is continuous at $\mathbb R$, then $F$ is differentiable by FTC at $\mathbb R$ and
$\forall X\in \mathbb R\;\; \; F'(X)=f(X)$
$g$ is differentiable at $\mathbb R$ then $G$ is dfferentiable at $\mathbb R$ and
$$\forall x\in \mathbb R\;\; G'(x)=F'(g(x))g'(x)=3x^2\cos(x^{12}).$$
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