Directly evaluate a definite integral of an explicitly given function, possibly involving symmetry properties, special techniques, or parameter determination to find a numerical or symbolic result.
We define on $[ 0,1 ]$ the function $P _ { n }$ by: $$\forall x \in [ 0,1 ] , \quad P _ { n } ( x ) = \frac { 1 } { n ! } \frac { \mathrm { d } ^ { n } \left( x ^ { n } ( 1 - x ) ^ { n } \right) } { \mathrm { d } x ^ { n } } .$$ We set $$P _ { n } ( x ) = \sum _ { k = 0 } ^ { n } a _ { k } x ^ { k }, \quad ( 1 - y ) ^ { n } = \sum _ { k = 0 } ^ { n } b _ { k } y ^ { k }$$ with for all $k \in \llbracket 0 , n \rrbracket , a _ { k } \in \mathbb { Z }$ and $b _ { k } \in \mathbb { Z }$. Let $n \in \mathbb { N } ^ { * }$. Justify the existence of $$I _ { n } = \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { 1 } \frac { ( 1 - y ) ^ { n } P _ { n } ( x ) } { 1 - x y } \mathrm {~d} x \mathrm {~d} y$$ and show that $$I _ { n } = \sum _ { \substack { r , s = 0 \\ r \neq s } } ^ { n } a _ { r } b _ { s } J _ { r , s } + \sum _ { r = 0 } ^ { n } a _ { r } b _ { r } J _ { r , r }$$
Problem 1: calculation of an integral For $x \geqslant 0$ we define $$f ( x ) = \int _ { 0 } ^ { \infty } \frac { e ^ { - t x } } { 1 + t ^ { 2 } } \mathrm {~d} t \quad \text { and } \quad g ( x ) = \int _ { 0 } ^ { \infty } \frac { \sin t } { t + x } \mathrm {~d} t$$ Deduce the value of $\int _ { 0 } ^ { \infty } \frac { \sin t } { t } \mathrm {~d} t$.
The equation $x^{2} + (b/a)x + (c/a) = 0$ has two real roots $\alpha$ and $\beta$. If $a > 0$, then the area under the curve $f(x) = x^{2} + (b/a)x + (c/a)$ between $\alpha$ and $\beta$ is (a) $(b^{2} - 4ac)/2a$ (b) $(b^{2} - 4ac)^{3/2}/6a^{3}$ (c) $-(b^{2} - 4ac)^{3/2}/6a^{3}$ (d) $-(b^{2} - 4ac)/2a$
Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f(0) = 0$, $f\left(\frac{\pi}{2}\right) = 3$ and $f'(0) = 1$. If $$g(x) = \int_x^{\frac{\pi}{2}} \left[f'(t)\operatorname{cosec} t - \cot t\operatorname{cosec} t\, f(t)\right] dt$$ for $x \in \left(0, \frac{\pi}{2}\right]$, then $\lim_{x \rightarrow 0} g(x) =$
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that its derivative $f'$ is continuous and $f(\pi) = -6$. If $F: [0, \pi] \rightarrow \mathbb{R}$ is defined by $F(x) = \int_{0}^{x} f(t)\, dt$, and if $$\int_{0}^{\pi} \left(f'(x) + F(x)\right) \cos x\, dx = 2$$ then the value of $f(0)$ is $\_\_\_\_$
Let $f : \left[ 0 , \frac { \pi } { 2 } \right] \rightarrow [ 0,1 ]$ be the function defined by $f ( x ) = \sin ^ { 2 } x$ and let $g : \left[ 0 , \frac { \pi } { 2 } \right] \rightarrow [ 0 , \infty )$ be the function defined by $g ( x ) = \sqrt { \frac { \pi x } { 2 } - x ^ { 2 } }$. The value of $2 \int _ { 0 } ^ { \frac { \pi } { 2 } } f ( x ) g ( x ) d x - \int _ { 0 } ^ { \frac { \pi } { 2 } } g ( x ) d x$ is $\_\_\_\_$ .
Let $f : \left[ 0 , \frac { \pi } { 2 } \right] \rightarrow [ 0,1 ]$ be the function defined by $f ( x ) = \sin ^ { 2 } x$ and let $g : \left[ 0 , \frac { \pi } { 2 } \right] \rightarrow [ 0 , \infty )$ be the function defined by $g ( x ) = \sqrt { \frac { \pi x } { 2 } - x ^ { 2 } }$. The value of $\frac { 16 } { \pi ^ { 3 } } \int _ { 0 } ^ { \frac { \pi } { 2 } } f ( x ) g ( x ) d x$ is $\_\_\_\_$ .
Let $f$ be a differentiable function from $R$ to $R$ such that $|f(x) - f(y)| \leq 2|x-y|^{3/2}$, for all $x,y \in R$. If $f(0) = 1$ then $\int_0^1 f^2(x)\,dx$ is equal to (1) 0 (2) 1 (3) 2 (4) $\frac{1}{2}$