Calculus Examples

Evaluate the Integral integral of x square root of 1-x^2 with respect to x
x1-x2dx
Step 1
Let u=1-x2. Then du=-2xdx, so -12du=xdx. Rewrite using u and du.
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Step 1.1
Let u=1-x2. Find dudx.
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Step 1.1.1
Differentiate 1-x2.
ddx[1-x2]
Step 1.1.2
Differentiate.
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Step 1.1.2.1
By the Sum Rule, the derivative of 1-x2 with respect to x is ddx[1]+ddx[-x2].
ddx[1]+ddx[-x2]
Step 1.1.2.2
Since 1 is constant with respect to x, the derivative of 1 with respect to x is 0.
0+ddx[-x2]
0+ddx[-x2]
Step 1.1.3
Evaluate ddx[-x2].
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Step 1.1.3.1
Since -1 is constant with respect to x, the derivative of -x2 with respect to x is -ddx[x2].
0-ddx[x2]
Step 1.1.3.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
0-(2x)
Step 1.1.3.3
Multiply 2 by -1.
0-2x
0-2x
Step 1.1.4
Subtract 2x from 0.
-2x
-2x
Step 1.2
Rewrite the problem using u and du.
u1-2du
u1-2du
Step 2
Simplify.
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Step 2.1
Move the negative in front of the fraction.
u(-12)du
Step 2.2
Combine u and 12.
-u2du
-u2du
Step 3
Since -1 is constant with respect to u, move -1 out of the integral.
-u2du
Step 4
Since 12 is constant with respect to u, move 12 out of the integral.
-(12udu)
Step 5
Use nax=axn to rewrite u as u12.
-12u12du
Step 6
By the Power Rule, the integral of u12 with respect to u is 23u32.
-12(23u32+C)
Step 7
Simplify.
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Step 7.1
Rewrite -12(23u32+C) as -1223u32+C.
-1223u32+C
Step 7.2
Simplify.
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Step 7.2.1
Multiply 23 by 12.
-232u32+C
Step 7.2.2
Multiply 3 by 2.
-26u32+C
Step 7.2.3
Cancel the common factor of 2 and 6.
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Step 7.2.3.1
Factor 2 out of 2.
-2(1)6u32+C
Step 7.2.3.2
Cancel the common factors.
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Step 7.2.3.2.1
Factor 2 out of 6.
-2123u32+C
Step 7.2.3.2.2
Cancel the common factor.
-2123u32+C
Step 7.2.3.2.3
Rewrite the expression.
-13u32+C
-13u32+C
-13u32+C
-13u32+C
-13u32+C
Step 8
Replace all occurrences of u with 1-x2.
-13(1-x2)32+C
 [x2  12  π  xdx ]