Indefinite Integration 1 Question 42
43. If $\int _1^{3} x^{2} F^{\prime}(x) d x=-12$ and $\int _1^{3} x^{3} F^{\prime \prime}(x) d x=40$, then the correct expression(s) is/are
(a) $9 f^{\prime}(3)+f^{\prime}(1)-32=0$
(b) $\int _1^{3} f(x) d x=12$
(c) $9 f^{\prime}(3)-f^{\prime}(1)+32=0$
(d) $\int _1^{3} f(x) d x=-12$
Passage II
For every function $f(x)$ which is twice differentiable, these will be good approximation of
$$ \int _a^{b} f(x) d x=\frac{b-a}{2}{f(a)+f(b)}, $$
for more acurate results for $c \in(a, b)$,
$$ F(c)=\frac{c-a}{2}[f(a)-f(c)]+\frac{b-c}{2}[f(b)-f(c)] $$
When $c=\frac{a+b}{2}$
$\int _a^{b} f(x) d x=\frac{b-a}{4}{f(a)+f(b)+2 f(c)} d x$
$(2006,6 M)$
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Solution:
- Given, $\int _1^{3} x^{2} F^{\prime}(x) d x=-12$
$\Rightarrow \quad\left[x^{2} F(x)\right] _1^{3}-\int _1^{3} 2 x \cdot F(x) d x=-12$
$\Rightarrow \quad 9 F(3)-F(1)-2 \int _1^{3} f(x) d x=-12$
$$ \begin{aligned} & \Rightarrow \quad-36-0-2 \int _1^{3} f(x) d x=-12 \\ & \therefore \quad \int _1^{3} f(x) d x=-12 \text { and } \int _1^{3} x^{3} F^{\prime \prime}(x) d x=40 \end{aligned} $$
$\Rightarrow \quad\left[x^{3} F^{\prime}(x)\right] _1^{3}-\int _1^{3} 3 x^{2} F^{\prime}(x) d x=40$
$\Rightarrow \quad\left[x^{2}\left(x F^{\prime}(x)\right] _1^{3}-3 \times(-12)=40\right.$
$\Rightarrow \quad{x^{2} \cdot\left[f^{\prime}(x)-F(x)\right] } _1^{3}=4$
$\Rightarrow \quad 9\left[f^{\prime}(3)-F(3)\right]-\left[f^{\prime}(1)-F(1)\right]=4$
$\Rightarrow \quad 9\left[f^{\prime}(3)+4\right]-\left[f^{\prime}(1)-0\right]=4$
$\Rightarrow \quad 9 f^{\prime}(3)-f^{\prime}(1)=-32$
