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# 𝘶-substitution with definite integrals

Performing u-substitution with definite integrals is very similar to how it's done with indefinite integrals, but with an added step: accounting for the limits of integration. Let's see what this means by finding integral, start subscript, 1, end subscript, squared, start color #7854ab, 2, x, end color #7854ab, start color #e07d10, left parenthesis, end color #e07d10, start color #1fab54, x, squared, plus, 1, end color #1fab54, start color #e07d10, right parenthesis, cubed, end color #e07d10, start color #7854ab, d, x, end color #7854ab.
We notice that start color #7854ab, 2, x, end color #7854ab is the derivative of start color #1fab54, x, squared, plus, 1, end color #1fab54, so u-substitution applies. Let start color #1fab54, u, equals, x, squared, plus, 1, end color #1fab54, then start color #7854ab, d, u, equals, 2, x, d, x, end color #7854ab. Now we substitute:
integral, start subscript, 1, end subscript, squared, start color #7854ab, 2, x, end color #7854ab, start color #e07d10, left parenthesis, end color #e07d10, start color #1fab54, x, squared, plus, 1, end color #1fab54, start color #e07d10, right parenthesis, cubed, end color #e07d10, start color #7854ab, d, x, end color #7854ab, equals, integral, start subscript, 1, end subscript, squared, start color #e07d10, left parenthesis, end color #e07d10, start color #1fab54, u, end color #1fab54, start color #e07d10, right parenthesis, cubed, end color #e07d10, start color #7854ab, d, u, end color #7854ab
Wait a minute! The limits of integration were fitted for x, not for u. Think about this graphically. We wanted the area under the curve start color #11accd, y, equals, 2, x, left parenthesis, x, squared, plus, 1, right parenthesis, cubed, end color #11accd between x, equals, 1 and x, equals, 2.
Function y = 2 x left parenthesis x squared + 1 right parenthesis cube is graphed. The x-axis goes from 0 to 3. The graph is a curve. The curve starts in quadrant 2, moves upward away from the x-axis to (2, 500). The region between the curve and the x-axis, between x = 1 and x = 2, is shaded.
Now that we changed the curve to start color #aa87ff, y, equals, u, cubed, end color #aa87ff, why should the limits stay the same?
Functions y = 2 x left parenthesis x squared + 1 right parenthesis cube and y = u cubed are graphed together. The graph of y = u cubed starts in quadrant 2, moves upward away from the x-axis and ends at about (3, 27).
Both start color #11accd, y, equals, 2, x, left parenthesis, x, squared, plus, 1, right parenthesis, cubed, end color #11accd and start color #aa87ff, y, equals, u, cubed, end color #aa87ff are graphed. You can see the areas under the curves between x, equals, 1 and x, equals, 2 (or u, equals, 1 and u, equals, 2) are very different in size.
Indeed, the limits shouldn't stay the same. To find the new limits, we need to find what values of start color #1fab54, u, end color #1fab54 correspond to start color #1fab54, x, squared, plus, 1, end color #1fab54 for x, equals, start color #ca337c, 1, end color #ca337c and x, equals, start color #ca337c, 2, end color #ca337c:
• Lower bound: left parenthesis, start color #ca337c, 1, end color #ca337c, right parenthesis, squared, plus, 1, equals, start color #ca337c, 2, end color #ca337c
• Upper bound: left parenthesis, start color #ca337c, 2, end color #ca337c, right parenthesis, squared, plus, 1, equals, start color #ca337c, 5, end color #ca337c
Now we can correctly perform the u-substitution:
integral, start subscript, start color #ca337c, 1, end color #ca337c, end subscript, start superscript, start color #ca337c, 2, end color #ca337c, end superscript, start color #7854ab, 2, x, end color #7854ab, start color #e07d10, left parenthesis, end color #e07d10, start color #1fab54, x, squared, plus, 1, end color #1fab54, start color #e07d10, right parenthesis, cubed, end color #e07d10, start color #7854ab, d, x, end color #7854ab, equals, integral, start subscript, start color #ca337c, 2, end color #ca337c, end subscript, start superscript, start color #ca337c, 5, end color #ca337c, end superscript, start color #e07d10, left parenthesis, end color #e07d10, start color #1fab54, u, end color #1fab54, start color #e07d10, right parenthesis, cubed, end color #e07d10, start color #7854ab, d, u, end color #7854ab
Functions y = 2 x left parenthesis x squared + 1 right parenthesis cube and y = u cubed are graphed together. The x-axis goes from negative 1 to 6. Each graph moves upward away from the x-axis. The first function ends at (2, 500). The region between the curve and the x-axis between x = 1 and x = 2 is shaded. The second function ends at about (6, 210). The region between the curve and the x-axis, between x = 1 and x = 5, is shaded. The 2 shaded regions look similar in size.
start color #aa87ff, y, equals, u, cubed, end color #aa87ff is graphed with the area from u, equals, 2 to u, equals, 5. Now we can see that the shaded areas look roughly the same size (they are actually exactly the same size, but it's hard to say by just looking).
From here on, we can solve everything according to u:
\begin{aligned} \displaystyle\int_{2}^5 u^3\,du&=\left[\dfrac{u^4}{4}\right]_{2}^5 \\\\ &=\dfrac{5^4}{4}-\dfrac{2^4}{4} \\\\ &=152.25 \end{aligned}
Remember: When using u-substitution with definite integrals, we must always account for the limits of integration.
Problem 1
Ella was asked to find integral, start subscript, 1, end subscript, start superscript, 5, end superscript, left parenthesis, 2, x, plus, 1, right parenthesis, left parenthesis, x, squared, plus, x, right parenthesis, cubed, d, x. This is her work:
Step 1: Let u, equals, x, squared, plus, x
Step 2: d, u, equals, left parenthesis, 2, x, plus, 1, right parenthesis, d, x
Step 3:
integral, start subscript, 1, end subscript, start superscript, 5, end superscript, left parenthesis, 2, x, plus, 1, right parenthesis, left parenthesis, x, squared, plus, x, right parenthesis, cubed, d, x, equals, integral, start subscript, 1, end subscript, start superscript, 5, end superscript, u, cubed, d, u
Step 4:
\begin{aligned} \displaystyle\int_1^5 u^3du&=\left[\dfrac{u^4}{4}\right]_1^5 \\\\ &=\dfrac{5^4}{4}-\dfrac{1^4}{4} \\\\ &=156 \end{aligned}
Is Ella's work correct? If not, what is her mistake?