Integration. No, it's differentiation!

Calculus Level 3

lim ⁡ x → 0 1 x ( ∫ y a e sin ⁡ 2 ( t ) d t − ∫ x + y a e sin ⁡ 2 ( t ) d t ) = ? \large \lim_{x \to 0} \frac 1x \left(\int_y^a e^{\sin^2(t)} dt - \int_{x+y}^a e^{\sin^2(t)} dt \right)= ?

e sin ⁡ 2 ( y ) \Large e^{\sin^2 (y)} e cos ⁡ ( 2 y ) \Large e^{\cos(2y)} e 2 sin ⁡ ( y ) \Large e^{2\sin(y)} e csc ⁡ 2 ( y ) \Large e^{\csc^2(y)}

Krishnendu Roy by krishnendu Roy , 19, India

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2 solutions

Chew-Seong Cheong
May 27, 2019

Similar solution with @Krishnendu Roy's

Let F ( x ) = ∫ 0 x e sin ⁡ 2 t d t \displaystyle F(x) = \int_0^x e^{\sin^2 t} dt . Then we have:

L = lim ⁡ x → 0 1 x ( ∫ y a e sin ⁡ 2 t d t − ∫ x + y a e sin ⁡ 2 t d t ) = lim ⁡ x → 0 1 x ( F ( a ) − F ( y ) − F ( a ) + F ( x + y ) ) Replace x with h = lim ⁡ h → 0 F ( y + h ) − F ( y ) h which is the definition of differentiation. = d F ( y ) d y = e sin ⁡ 2 y \begin{aligned} L & = \lim_{x \to 0} \frac 1x \left(\int_y^a e^{\sin^2 t} dt - \int_{x+y}^a e^{\sin^2 t} dt \right) \\ & = \lim_{x \to 0} \frac 1x \left(F(a) - F(y) -F(a) + F(x+y) \right) & \small \color{#3D99F6} \text{Replace }x \text{ with }h \\ & = \lim_{h \to 0} \frac {F(y+h) - F(y)}h & \small \color{#3D99F6} \text{which is the definition of differentiation.} \\ & = \frac {d F(y)}{dy} \\ & = \boxed{e^{\sin^2 y}} \end{aligned}

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Krishnendu Roy
May 27, 2019

after simplifying above equation we get

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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