Question

19
if $f(x) = \int_{x}^{x^4} t^3 dt$
then
$f(x) = $
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question 5
if $d(v) = \int_{4v}^{7v} \frac{t^2}{t^2 + 16} dt$ then
$d(v) = $

Explanation:

Step1: Apply FTC and chain rule (Q1)

Recall the Fundamental Theorem of Calculus (FTC) and chain rule: if $f(x)=\int_{a(x)}^{b(x)} g(t)dt$, then $f'(x)=g(b(x))\cdot b'(x) - g(a(x))\cdot a'(x)$.
For $f(x)=\int_{x}^{x^3} t^3 dt$, $g(t)=t^3$, $a(x)=x$, $b(x)=x^3$.
Compute $g(b(x))=(x^3)^3=x^9$, $b'(x)=3x^2$; $g(a(x))=x^3$, $a'(x)=1$.
So $f'(x)=x^9\cdot 3x^2 - x^3\cdot 1$

Step2: Simplify the expression (Q1)

$$\begin{align} f'(x)&=3x^{11}-x^3 \end{align}$$

Step3: Apply FTC and chain rule (Q2)

For $d(v)=\int_{4v}^{7v} \frac{t^2}{t^2+16} dt$, $g(t)=\frac{t^2}{t^2+16}$, $a(v)=4v$, $b(v)=7v$.
Compute $g(b(v))=\frac{(7v)^2}{(7v)^2+16}=\frac{49v^2}{49v^2+16}$, $b'(v)=7$; $g(a(v))=\frac{(4v)^2}{(4v)^2+16}=\frac{16v^2}{16v^2+16}=\frac{v^2}{v^2+1}$, $a'(v)=4$.
So $d'(v)=\frac{49v^2}{49v^2+16}\cdot7 - \frac{v^2}{v^2+1}\cdot4$

Step4: Simplify the expression (Q2)

$$\begin{align} d'(v)&=\frac{343v^2}{49v^2+16}-\frac{4v^2}{v^2+1} \end{align}$$

Answer:

For Question 1: $3x^{11}-x^3$
For Question 2: $\frac{343v^2}{49v^2+16}-\frac{4v^2}{v^2+1}$