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FTC with a variable limit and a scaled improper integral

Math I: Calculus · Week 7 test, open question · 7 points · Lectures 11–12

Problem

This question consists of two independent parts.

Part A. Define

F(x)=∫0x2t2e−t dt.F(x)=\int_0^{x^2}t^2\mathrm{e}^{-t}\,\mathrm{d}t.

Find F′(x).F'(x)\text{.} [2 points]

Part B. Let a>0a>0 and define

I(a)=∫0∞x2e−ax dx.I(a)=\int_0^{\infty}x^2\mathrm{e}^{-ax}\,\mathrm{d}x.
  • (B1) Apply the substitution t=axt=ax first on [0,R],[0,R]\text{,} then let R→∞,R\to\infty\text{,} to prove that I(a)=I(1)/a3.I(a)=I(1)/a^3\text{.} Justify the transformed upper limit. [2 points]
  • (B2) Evaluate I(1)I(1) by integration by parts and hence find I(a).I(a)\text{.} Justify every boundary limit that occurs. [3 points]

What you need

Plan

Part A is a composition: an integral with a variable upper limit, evaluated at x2.x^2\text{.} Part B is about doing improper integrals honestly. Every substitution and every integration by parts happens on a bounded interval [0,R];[0,R]\text{;} only then do you let R→∞.R\to\infty\text{.} The gamma function is the general version of I(1).I(1)\text{.}

Step-by-step solution

Try each step yourself before you open it.

Part A · FTC and the chain rule (2 points)

The integrand f(t)=t2e−tf(t)=t^2\mathrm{e}^{-t} is continuous, so by FTC part 1 the function

G(u)=∫0ut2e−t dtG(u)=\int_0^u t^2\mathrm{e}^{-t}\,\mathrm{d}t

is differentiable with G′(u)=u2e−u.G'(u)=u^2\mathrm{e}^{-u}\text{.} Since F(x)=G(x2),F(x)=G(x^2)\text{,} the chain rule gives

F′(x)=G′(x2)⋅2x=2x (x2)2e−x2=2x5e−x2.F'(x)=G'(x^2)\cdot2x=2x\,(x^2)^2\mathrm{e}^{-x^2}=2x^5\mathrm{e}^{-x^2}.

Concept: Fundamental theorem of calculus, part 1

(B1) Step 1 · Substitute on [0, R] (1 point)

For R>0,R>0\text{,} let t=ax.t=ax\text{.} Then x=t/ax=t/a and dx=dt/a;\mathrm{d}x=\mathrm{d}t/a\text{;} the limits x=0x=0 and x=Rx=R become t=0t=0 and t=aR.t=aR\text{.} So

∫0Rx2e−ax dx=1a3∫0aRt2e−t dt.\int_0^R x^2\mathrm{e}^{-ax}\,\mathrm{d}x=\frac{1}{a^3}\int_0^{aR}t^2\mathrm{e}^{-t}\,\mathrm{d}t.

Concept: Integration by substitution

(B1) Step 2 · Let R → ∞ (1 point)

Because a>0,a>0\text{,} aR→∞aR\to\infty as R→∞.R\to\infty\text{.} This is the justification the question asks for: if aa were negative, the upper limit would go to −∞-\infty instead. Taking limits on both sides,

I(a)=1a3lim⁡S→∞∫0St2e−t dt=I(1)a3,I(a)=\frac{1}{a^3}\lim_{S\to\infty}\int_0^{S}t^2\mathrm{e}^{-t}\,\mathrm{d}t=\frac{I(1)}{a^3},

where the limit on the right exists because I(1)I(1) converges (B2 computes it).

Concept: Improper integral

(B2) Step 1 · Integrate by parts twice (2 points)

On [0,R],[0,R]\text{,} with u=x2u=x^2 and v′=e−x,v'=\mathrm{e}^{-x}\text{,}

∫0Rx2e−x dx=[−x2e−x]0R+2∫0Rxe−x dx,∫0Rxe−x dx=[−xe−x]0R+∫0Re−x dx.\begin{aligned} \int_0^R x^2\mathrm{e}^{-x}\,\mathrm{d}x &=\left[-x^2\mathrm{e}^{-x}\right]_0^R +2\int_0^R x\mathrm{e}^{-x}\,\mathrm{d}x,\\ \int_0^R x\mathrm{e}^{-x}\,\mathrm{d}x &=\left[-x\mathrm{e}^{-x}\right]_0^R +\int_0^R\mathrm{e}^{-x}\,\mathrm{d}x. \end{aligned}

Concept: Integration by parts

(B2) Step 2 · Boundary terms and the answer (1 point)

The boundary terms are −R2e−R-R^2\mathrm{e}^{-R} and −Re−R,-R\mathrm{e}^{-R}\text{,} and both tend to 00 as R→∞,R\to\infty\text{,} because the exponential beats any power. At x=0x=0 both are 0.0\text{.} Also ∫0Re−x dx=1−e−R→1.\int_0^R\mathrm{e}^{-x}\,\mathrm{d}x=1-\mathrm{e}^{-R}\to1\text{.} Hence

I(1)=2(0+1)=2,I(a)=2a3.I(1)=2\left(0+1\right)=2,\quad I(a)=\frac{2}{a^3}.

Concept: Limit at infinity

Where marks are lost

  • Forgetting the factor 2x2x in Part A. The upper limit is x2;x^2\text{;} FTC alone gives G′(u),G'(u)\text{,} and the chain rule supplies 2x.2x\text{.}
  • Substituting directly on [0,∞).[0,\infty)\text{.} The question asks for [0,R][0,R] first, and the mark is for saying why aR→∞.aR\to\infty\text{.}
  • Writing [−x2e−x]0∞=0[-x^2\mathrm{e}^{-x}]_0^{\infty}=0 without a reason. Each boundary limit needs its own sentence.

Try next

Symbols: gamma, infinity, integral