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Given that $$\int_{c}^xf(t)\,\mathrm{d}t=x^3 + x^5,$$ where $c$ is constant, find the value of $c$.

I started by getting finding $D_x(\int_{c}^xf(t)\,\mathrm{d}t)$:

$$D_x\left(\int_{c}^xf(t)\,\mathrm{d}t\right)=D_x(x^3 + x^5) = 3x^2 + 5x^4.$$

Thus we can say that $f(x)$ $=$ $3x^2$ $+$ $5x^4$.

I am stuck after this step. Kindly help me. Thanks in advance.

Ѕᴀᴀᴅ
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PRD
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  • Hint: it’s zero. – Jacky Chong Feb 22 '18 at 02:12
  • Thank you @JackyChong. The idea the c=0 also came to my mind. I want to know how to show it algebraically. Kindly help me if you have an idea how to do show it. Thank you! – PRD Feb 22 '18 at 02:16
  • If $x=0$ then the RHS evaluates to zero. In the LHS the integral will evaluate to zero only if $c=0$. – 3.14159 Feb 22 '18 at 06:23

1 Answers1

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Realize that if $c \neq 0$, there would be some constant part, like so:

$$\int_{c}^x f(t)dt = x^3 + x^5 - c^3 - c^5$$

But since there is no constant part, you can deduce that $c = 0$. You can show a bit clearer, applying the Fundamental Theorem of Calculus as you did:

$$ \require{cancel} \dfrac{d}{dx}\int_c^x f(t)dt = \dfrac{d}{dx}\left(x^3+x^5\right) \Rightarrow f(t) = 3t^2 + 5t^4$$ $$\int_c^x \left(3t^2 + 5x^4\right) dt = \left(x^3 + 5x^4\right) - \left(c^3 + c^5\right)$$ $$x ^3 + x^5 = \left(x^3 + x^5\right) - \left(c^3 + c^5\right)$$ $$0 = - \left(c^3 + c^5\right)$$ $$0 = -c^3\cancel{\left(1+c^2\right)}$$ $$c = 0$$

Andrew Li
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