Suppose we define the definite integral using the following formula $$\int_\limits{a}^{b} f(x) d x=\frac{b-a}{2}(f(a)+f(b))$$, for more accurate result for
$$c \in(a, b) \mathrm{F}(c)=\frac{c-a}{2}(f(a)+f(c))+\frac{b-c}{2}(f(b)+f(c))$$.
When $$c=\frac{a+b}{c}, \int_\limits{a}^{b} f(x) d x=\frac{b-a}{4}(f(a)+f(b)+2 f(c))$$
$$\int_\limits{0}^{\pi / 2} \sin x d x$$ is equal to:
Suppose we define the definite integral using the following formula $$\int_\limits{a}^{b} f(x) d x=\frac{b-a}{2}(f(a)+f(b))$$, for more accurate result for
$$c \in(a, b) \mathrm{F}(c)=\frac{c-a}{2}(f(a)+f(c))+\frac{b-c}{2}(f(b)+f(c))$$.
When $$c=\frac{a+b}{c}, \int_\limits{a}^{b} f(x) d x=\frac{b-a}{4}(f(a)+f(b)+2 f(c))$$
If $$\lim_\limits{t \rightarrow a} \frac{\int_{a}^{t} f(x) d x-\frac{(t-a)}{2}\{f(t)+f(a)\}}{(t-a)^{3}}=0$$ then the degree of polynomial function $$f(x)$$ almost is:
Suppose we define the definite integral using the following formula $$\int_\limits{a}^{b} f(x) d x=\frac{b-a}{2}(f(a)+f(b))$$, for more accurate result for
$$c \in(a, b) \mathrm{F}(c)=\frac{c-a}{2}(f(a)+f(c))+\frac{b-c}{2}(f(b)+f(c))$$.
When $$c=\frac{a+b}{c}, \int_\limits{a}^{b} f(x) d x=\frac{b-a}{4}(f(a)+f(b)+2 f(c))$$
$$f''(x) < 0 \forall x \in(a, b)$$ and $$c$$ is a point such that $$a < c < b$$, and $$(c, f(C))$$ is the point lying on the curve for which $$\mathrm{F}(C)$$ is maximum, then $$f'(C)$$ is equal to:
$$ \text { Match the following : } $$
| (i) | $$ \int_0^{\pi / 2}(\sin x)^{\cos x}\left(\cos x \cot x-\log \left(\sin ^x\right)^{\sin } x\right) \mathrm{d} x $$ |
(A) | 1 |
|---|---|---|---|
| (ii) | $$ \text { Area bounded by }-4 y^2=x \text { and } x-1=-5 y^2 $$ |
(B) | 0 |
| (iii) | Cosine of the angle of intersection of $y=3^{x-1} \log x$ and $y=x^{x-1}$ is | (C) | 6 In 2 |
| (iv) | $$ \frac{d y}{d x}=\frac{2}{(x+y)} ; y\left(-\frac{2}{3}\right)=0 \text {, then value of constant }(\mathrm{k})= $$ |
(D) | 4/3 |
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