f(x)=∫₀ˣ³sin(t)dt\nf(x)=□
f(x)=∫₀ˣ³sin(t)dt\nf(x)=□
Answer
Explanation:
Step1: Apply the chain - rule and FTC
By the fundamental theorem of calculus (FTC) and the chain - rule, if $F(x)=\int_{a}^{u(x)}f(t)dt$, then $F'(x)=f(u(x))\cdot u'(x)$. Here, $a = 0$, $u(x)=x^{3}$, and $f(t)=\sin(t)$.
Step2: Find $u'(x)$
Differentiate $u(x)=x^{3}$ with respect to $x$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $u'(x)=3x^{2}$.
Step3: Calculate $F'(x)$
Since $f(u(x))=\sin(u(x))=\sin(x^{3})$ and $u'(x)=3x^{2}$, then $F'(x)=\sin(x^{3})\cdot3x^{2}=3x^{2}\sin(x^{3})$.
Answer:
$3x^{2}\sin(x^{3})$