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myrzilka [38]
3 years ago
14

How do you do this problem?

Mathematics
2 answers:
irga5000 [103]3 years ago
7 0

Answer:

\int\limits^{\sqrt{\pi}}_0 {4x^3\cos(x^2)} \, dx=-4

Step-by-step explanation:

So we have the integral:

\int\limits^{\sqrt{\pi}}_0 {4x^3\cos(x^2)} \, dx

As told, let's use u-substitution first and then use integration by parts.

For the u-substitution, we can let u to be equal to x². So:

u=x^2

Differentiate:

du=2x\, dx

We can rewrite our integral as:

\int\limits^{\sqrt{\pi}}_0 {2x(2x^2)\cos(x^2)} \, dx

Therefore, by making our u-substitution, our integral is now:

\int\limits {2u\cos(u)} \, du

We also need to change our bounds. Substitute them into u. So:

u=\sqrt{\pi}^2=\pi\\u=(0)^2=0

Therefore, our integral with our new bounds is:

\int\limits^{\pi}_{0} {2u\cos(u)} \, du

Now, let's use integration by parts. Integration by parts is given by:

\int\limits {v}\, dy=vy-\int y\, dv

(I changed the standard u to y because we are already using u).

Let's let v be 2u and let's let dy be cos(u). Thus:

v=2u\\dv=2\,du

And:

dy=\cos(u)\\y=\sin(u)

So, do integration by parts:

=2u\sin(u)-\int \sin(u)2\,du

Simplify:

=2u\sin(u)-2\int \sin(u)\,du

Evaluate the integral:

=2u\sin(u)+2\cos(u)

Now, use the bounds. So:

(2(\pi)\sin(\pi)+2\cos(\pi))-(2(0)\sin(0)+2\cos(0))

Evaluate:

=(2\pi(0)+2(-1))-(0(0)+2(1))

Simplify:

=(-2)-(2)

Subtract:

=-4

And we're done!

Vadim26 [7]3 years ago
3 0

Answer:

Your answer is absolutely correct

Step-by-step explanation:

The work would be as follows:

\int _0^{\sqrt{\pi }}4x^3\cos \left(x^2\right)dx,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=> 4\cdot \int _0^{\sqrt{\pi }}x^3\cos \left(x^2\right)dx\\\\\mathrm{Apply\:u-substitution:}\:u=x^2\\=> 4\cdot \int _0^{\pi }\frac{u\cos \left(u\right)}{2}du\\\\\mathrm{Apply\:Integration\:By\:Parts:}\:u=u,\:v'=\cos \left(u\right)\\=> 4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\int \sin \left(u\right)du\right]^{\pi }_0\\\\

\int \sin \left(u\right)du=-\cos \left(u\right)\\=> 4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\left(-\cos \left(u\right)\right)\right]^{\pi }_0\\\\\mathrm{Simplify\:}4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\left(-\cos \left(u\right)\right)\right]^{\pi }_0:\quad 2\left[u\sin \left(u\right)+\cos \left(u\right)\right]^{\pi }_0\\\\\mathrm{Compute\:the\:boundaries}:\quad \left[u\sin \left(u\right)+\cos \left(u\right)\right]^{\pi }_0=-2\\=> 2(-2) = - 4

Hence proved that your solution is accurate.

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