170- Quiz 11

The 11th quiz is here.

Solutions follow.

Problem 1. Find the derivative of \displaystyle G(x)=\int_1^{5x^2}(t-\sin(t))\,dt.

There are two ways to proceed. We can solve this problem by direct calculation: First we compute an antiderivative H(t) of t-\sin t, then use it to compute G(x)=H(5x^2)-H(1), and finally use this expression to compute G'(x).

We have \displaystyle \int (t-\sin t)\,dt=\frac{t^2}2+\cos t+C, so \displaystyle G(x)=\left.\frac{t^2}2+\cos t\right|_1^{5x^2}=\frac{(5x^2)^2}2+\cos(5x^2)-\frac12-\cos1, and

\displaystyle G'(x)=\frac{2(5x^2)10x}2-\sin(5x^2)10x \displaystyle =(5x^2-\sin(5x^2))10x.

The second method uses instead the fundamental theorem of calculus directly, and avoids having to compute any antiderivatives: Suppose f is a continuous function on the interval {}[a,b] and that H is an antiderivative of f. Recall that this simply means that H'(t)=f(t) for a<t<b.

Suppose g is a differentiable function, and G is defined by \displaystyle G(x)=\int_a^{g(x)}f(t)\,dt. By the fundamental theorem of calculus, this means that \displaystyle G(x)=H(g(x))-H(a), and therefore

\displaystyle G'(x)=H'(g(x))\cdot g'(x)=f(g(x))\cdot g'(x),

using the chain rule. In the particular case that we are interested in, a=1, g(x)=5x^2, and f(t)=t-\sin t, so

\displaystyle G'(x)=(5x^2-\sin(5x^2))10x,

as before.

Problem 2. Find \displaystyle \int_0^{\sqrt \pi/2} x\sec^2(x^2)\tan(x^2)\,dx.

We use the substitution method to simplify the integrand. Although this can be done directly, I will do it in two stages for clarity. First, the most obvious difficulty with the expression at hand is the x^2 that appears as the argument in the trigonometric functions. This suggests to first try the substitution

t=x^2.

Note that dt=2x\,dx, and the integrand already contains a factor of x, which is a good sign. To rewrite the integral in terms of t, we note that if x=0, then t=0^2=0, while if x=\sqrt\pi/2, then t=(\sqrt\pi/2)^2=\pi/4, so the integral becomes

\displaystyle \int_0^{\pi/4}\frac12\sec^2(t)\tan(t)\,dt.

This expression is still complicated. There are two ways to proceed now. We may use that (\tan t)'=\sec^2 t. This suggests the substitution u=\tan(t). Then du=\sec^2(t)\,dt. When t=0, we have u=\tan(0)=0, while when t=\pi/4, we have u=\tan(\pi/4)=1, so the integral becomes

\displaystyle \int_0^1 \frac12u\,du=\left.\frac{u^2}4\right|_0^1=\frac14.

The second way to proceed is to use that (\sec t)'=(\sec t)(\tan t). This suggests the substitution w=\sec(t). Then dw=\sec(t)\tan(t)\,dt. When t=0, we have w=\sec(0)=1, while when t=\pi/4, we have w=\sec(\pi/4)=\sqrt2, so the integral becomes

\displaystyle \int_1^{\sqrt 2}\frac12w\,dw=\left.\frac{w^2}4\right|_1^{\sqrt2}=\frac24-\frac14=\frac14,

as before.

I saw a slightly different presentation of the last approach in some of the solutions: Use the substitution y=\cos(x^2). Then dy=-2x\sin(x^2)\,dx. If x=0, then y=1, while if x=\sqrt\pi/2, then y=\cos(\pi/4)=1/\sqrt2. Then the integral becomes

\displaystyle \int_1^{1/\sqrt2}-\frac12\frac1{y^3}\,dy=\left.\frac{y^{-2}}4\right|_1^{1/\sqrt2}=\frac24-\frac14=\frac14.

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This entry was posted on Friday, December 10th, 2010 at 11:42 am and is filed under 170: Calculus I. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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