If $f(x)=\displaystyle\int_1^x \dfrac{e^{\sin u}}{u}\,du,\ x1$ and $\displaystyle\int_1^3 \dfrac{e^{\sin x^2}}{x}\,dx=\dfrac12\big[f(a)-f(1)\big]$, then one of the possible value of $a$ is

(1) $3$ (2) $6$ (3) $9$ (option 4 is misprinted with no value in the source textboo

Question

If $f(x)=\displaystyle\int_1^x \dfrac{e^{\sin u}}{u}\,du,\ x>1$ and $\displaystyle\int_1^3 \dfrac{e^{\sin x^2}}{x}\,dx=\dfrac12\big[f(a)-f(1)\big]$, then one of the possible value of $a$ is

(1) $3$   (2) $6$   (3) $9$   (option 4 is misprinted with no value in the source textbook)

Accepted Answer

> [!TLDR]
>
> Substituting $u=x^2$ turns the given integral into exactly the form $\frac12[f(u)]_1^9$ using the defined function $f$, and matching this against the stated RHS $\frac12[f(a)-f(1)]$ identifies $a$ directly.

**Step 1. Set up the substitution $u=x^2$.** Then $du=2x\,dx$, so $dx=\dfrac{du}{2x}=\dfrac{du}{2\sqrt u}$ (taking $x=\sqrt u>0$ on $[1,3]$).

**Step 2. Rewrite the integrand in terms of $u$.**
$$\frac{e^{\sin x^2}}{x}\,dx=\frac{e^{\sin u}}{\sqrt u}\cdot\frac{du}{2\sqrt u}=\frac{e^{\sin u}}{2u}\,du.$$

**Step 3. Convert the limits.** When $x=1$, $u=1^2=1$. When $x=3$, $u=3^2=9$.

**Step 4. Rewrite the whole integral.**
$$\int_1^3\frac{e^{\sin x^2}}{x}\,dx=\int_1^9\frac{e^{\sin u}}{2u}\,du=\frac12\int_1^9\frac{e^{\sin u}}{u}\,du.$$

**Step 5. Express in terms of the given $f$.** Since $f(x)=\displaystyle\int_1^x\frac{e^{\sin u}}{u}\,du$, we have $\displaystyle\int_1^9\frac{e^{\sin u}}{u}\,du=f(9)-f(1)$. So …

Q.If $f(x)=\displaystyle\int_1^x \dfrac{e^{\sin u}}{u}\,du,\ x>1$ and $\displaystyle\int_1^3 \dfrac{e^{\sin x^2}}{x}\,dx=\dfrac12\big[f(a)-f(1)\big]$ , then one of the possible value of $a$ is

Concept understanding — Fundamental Theorem of Integral Calculus

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