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X
- WE WANT TO EVALUATE THIS DEFINITE INTEGRAL
USING INTEGRATION BY SUBSTITUTION.
SO FOR THE FIRST STEP LET'S GO AHEAD AND REWRITE THIS
AS A RATIONAL EXPONENT.
SO WE'LL HAVE THE INTEGRAL FROM 1 TO 2 OF 1
DIVIDED BY THE QUANTITY 7X + 2 RAISED TO THE 1/2 POWER DX.
SO HERE WE HAVE A COMPOSITE FUNCTION
SO WE'LL LET U EQUAL THE INNER FUNCTION OF 7X + 2.
SO DIFFERENTIAL U WILL BE 7DX,
BUT NOTICE HOW WE DON'T HAVE A 7DX AS PART OF OUR INTEGRAL,
WE JUST HAVE 1DX.
SO LET'S GO AHEAD AND DIVIDE BOTH SIDES BY SEVEN HERE.
SO WE HAVE 1/7DU = DX.
SO WE'RE GOING TO REPLACE 7X + 2 WITH THE U,
BUT NOW WE CAN REPLACE DX WITH 1/7DU.
SO LET'S GO AHEAD AND REWRITE THIS IN TERMS OF U.
NOW WE DON'T WANT TO INCLUDE THE LIMITS OF INTEGRATION ON HERE
BECAUSE THESE LIMITS OF INTEGRATION ARE IN TERMS OF X.
SO WE'LL HAVE THE INTEGRAL, THIS WILL BE 1/U TO THE 1/2,
WHICH I'M GOING TO WRITE AS U TO THE -1/2 AND THEN DX + 1/7DU.
SO WE'LL HAVE 1/7 HERE AND DU HERE
AND NOW WE'LL DETERMINE THE ANTI-DERIVATIVE OF U TO THE -1/2
WITH RESPECTS TO U.
SO WE'LL HAVE 1/7 x U TO THE -1/2 + 1,
THAT'S GOING TO BE U TO THE 1/2 DIVIDED BY 1/2.
LET'S GO AHEAD AND CLEAN THIS UP AND WRITE IT IN TERMS OF X.
INSTEAD OF DIVIDING BY 1/2 WE CAN MULTIPLY BY 2.
SO WE'LL HAVE 2/7 AND THEN U IS 7X + 2,
BUT REMEMBER THIS IS A DEFINITE INTEGRAL.
SO NOW WE NEED TO EVALUATE THIS AT TWO AND ONE.
SO WE'RE GOING TO HAVE 2/7 AND THEN WHEN X IS 2
WE'RE GOING TO HAVE 7 x 2 = 14 + 2 = 16 TO THE 1/2 POWER
OR THE SQUARE ROOT OF 16 - 1X IS 1.
WE'LL HAVE 7 x 1 = 7 + 2 - 9 TO THE 1/2.
SO WE'LL HAVE 2/7 x 4 - 3
SO THIS DEFINITE INTEGRAL IS EQUAL TO 2/7.
OKAY, I HOPE YOU FOUND THIS HELPFUL.