A calculus problem by Jose Sacramento

Calculus Level 4

Let f : [ − 1 , 0 ] → R f: [-1,0] \to \mathbb R be a function differentiable within the domain and that ∫ − 1 0 ( f ( x ) ) 2 d x = 10 \displaystyle \int_{-1}^0 \left(f(x)\right)^2 dx = 10 and f ( − 1 ) = 2 f(-1) = 2 . Find the value of the integral below.

∫ − 1 0 x f ′ ( x ) f ( x ) d x \large \int_{-1}^0 xf'(x)f(x) \ dx


The answer is -3.

Jose Sacramento by Jose Sacramento , 66, Portugal

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1 solution

∫ − 1 0 ( f ( x ) ) 2 d x = 10 By integration by parts. x ( f ( x ) ) 2 ∣ − 1 0 − 2 ∫ − 1 0 x f ′ ( x ) f ( x ) d x = 10 ( 0 ) ( f ( 0 ) ) 2 − ( − 1 ) ( f ( − 1 ) ) 2 − 2 ∫ − 1 0 x f ′ ( x ) f ( x ) d x = 10 0 + 4 − 2 ∫ − 1 0 x f ′ ( x ) f ( x ) d x = 10 ⟹ ∫ − 1 0 x f ′ ( x ) f ( x ) d x = − 3 \begin{aligned} \color{#3D99F6} \int_{-1}^0 \left(f(x)\right)^2 dx & =10 & \small \color{#3D99F6} \text{By integration by parts.} \\ \color{#3D99F6} x\left(f(x)\right)^2 \bigg|_{-1}^0 - 2 \int_{-1}^0 x f'(x) f(x)\ dx & = 10 \\ (0)\left(f(0)\right)^2 - (-1)\left(f(-1)\right)^2 - 2 \int_{-1}^0 x f'(x) f(x)\ dx & = 10 \\ 0 +4 - 2 \int_{-1}^0 x f'(x) f(x)\ dx & = 10 \\ \implies \int_{-1}^0 x f'(x) f(x)\ dx & = \boxed{-3} \end{aligned}

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