00:01
Hi, from the question given that we need to evaluate the integral minus 1 to plus 1 3 x square minus 2 dx.
00:13
So, here it is given that this is an even function.
00:16
So, if it is an even function integral minus a to plus a f of x dx can be written as 2 times of integral 0 to a f of x dx.
00:28
So, by using the same procedure integral 0 to 1 2 times of integral 0 to 1 3 x square minus 2 dx.
00:38
So, that is equal to 2 times of and we integrating this we obtain 3 x cube by 3 minus 2 x with the limit 0 to 1.
00:46
So, this 3 and this 3 will get cancelled.
00:48
So, we have 2 times of 1 cube is 1 minus 2 when we apply 0 then the whole term became vanishes.
00:54
So, that is equal to minus 2.
00:58
Hence, we conclude that simplification of the given expression minus 1 to plus 1 3 x square minus 2 dx is equal to minus 2.
01:08
Now, let us move on to the part b.
01:11
So, in part b we need to evaluate the integral sin 3 x cos 2 x dx.
01:19
So, in general we know that 2 times of sin a cos b is equal to sin a plus b plus sin a minus b.
01:34
So, on comparing this a is equal to 3 x and b is equal to 2 x.
01:40
Therefore, the above integral became integral 2 times of not 2 times of this is 1 by 2, 1 by 2 times of sin of a plus b is 3 x plus 2 x is 5 x plus sin of a minus b is 3 x minus 2 x is x dx.
02:01
So, 1 by 2 is common integral sin 5 x dx plus 1 by 2 integral sin x dx.
02:11
So, after integrating we obtain 1 over 2 integration of sin is minus cos 5 x divided by 5 minus cos x divided by 1 plus the integrating constant c.
02:29
So, that the final answer will be minus 1 by 10 cos 5 x minus cos x plus c.
02:40
Now, let us move on to the third one.
02:43
So, in part c we need to find the antiderivative of f of x is equal to 6 x to the power of 5 and that satisfies the given condition f of 1 is equal to 0.
02:53
So, antiderivative of f of x is equal to integral f of x dx...