Question

8 mark for review $\frac{d}{dx}left(int_{1}^{x}t^{4}+t^{2}+1dt
ight)=$
(a) $x^{4}+x^{2}+1$
(b) $4x^{3}+2x$
(c) $x^{4}+x^{2}-2$
(d) $4x^{3}+2x - 6$

Explanation:

Step1: Apply fundamental theorem of calculus

The fundamental theorem of calculus states that if $F(x)=\int_{a}^{x}f(t)dt$, then $F^\prime(x) = f(x)$. Here, $a = 1$ and $f(t)=t^{4}+t^{2}+1$.

Step2: Identify the result

Differentiating $\int_{1}^{x}t^{4}+t^{2}+1dt$ with respect to $x$ gives $x^{4}+x^{2}+1$.

Answer:

A. $x^{4}+x^{2}+1$