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Stumped by this one and I've got to get through fifty questions tonight. I tried to take some integrals, but it's obviously not as simple as I thought. Any hints/help/explanations would be greatly appreciated!

If $\int_0^3 e^{\sin x}dx=k$ , then $\int_1^2 xe^{\sin (4-x^2)}dx=$

(a) -k/2 ------------ (b) -k/3 ----------- (c) k/6

(d) k/3 --------------- (e) k/2

  • 2
    have you tried substitution? $z=4-x^2$ seems fairly natural... – 2017-01-24

2 Answers 2

1

Put $4-x^2 = t$ .....(1)

$-2xdx = dt$

$xdx = -\frac 12 dt$

Now when lower bound x=1 then from equation (1),

t = 3

Now when upper bound x=2 then from equation (1),

t = 0.

So our integral,

$\int_1^2 xe^{\sin (4-x^2)}dx= \int_3^0 e^{\sin (t)} (- \frac 12 dt )$

= $\frac k2$

  • 0
    Thanks! Would I be correct in saying that $\int_3^0 e^{\sin (t)} (- \frac 12 dt )$ is equal to $\int_0^3 e^{\sin (t)} (\frac 12 dt )$? – 2017-01-24
  • 1
    Yes you can. Its rule. – 2017-01-24
3

Let $u=4-x^2$ so that $du=-2xdx\implies x dx = -\frac 12 du$

$$\int_1^2 xe^{\sin (4-x^2)}dx= \int_3^0 e^{\sin (u)} (-\frac 12 du )=\frac k2$$

  • 0
    Totally makes sense, although I'm a little confused about what happens to the bounds of integration. I see that if you plug in 2 and 1 for $4-x^2$, you get the bounds 3 to 0. But integral questions in this form being new to me, it's all a bit confusing. – 2017-01-24
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    Additionally, it wasn't substituting that I found difficult, I just didn't quite see how getting $-1/2\int_1^2 -2xe^{sin(4-x^2)}dx$ showed a relationship with k. – 2017-01-24