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Math  /  Calculus

Question∫0−1(1−t2)3\int_{0}^{-1}\left(1-t^{2}\right)^{3}

Studdy Solution

STEP 1

What is this asking? We need to find the definite integral of a polynomial from 0 to -1. Watch out! Don't forget to expand the polynomial before integrating, and be careful with the negative limit of integration!

STEP 2

1. Expand the integrand
2. Integrate the expanded polynomial
3. Evaluate the definite integral

STEP 3

Alright, let's **expand** that expression inside the integral!
We have (1−t2)3(1-t^2)^3.
Remember, this means (1−t2)⋅(1−t2)⋅(1−t2)(1-t^2) \cdot (1-t^2) \cdot (1-t^2).

STEP 4

First, let's multiply the first two terms: (1−t2)⋅(1−t2)=1−t2−t2+t4=1−2t2+t4(1-t^2) \cdot (1-t^2) = 1 - t^2 - t^2 + t^4 = 1 - 2t^2 + t^4

STEP 5

Now, multiply this result by the third term: (1−2t2+t4)(1−t2)=1−t2−2t2+2t4+t4−t6=1−3t2+3t4−t6(1 - 2t^2 + t^4)(1-t^2) = 1 - t^2 - 2t^2 + 2t^4 + t^4 - t^6 = 1 - 3t^2 + 3t^4 - t^6 So, (1−t2)3=1−3t2+3t4−t6(1-t^2)^3 = 1 - 3t^2 + 3t^4 - t^6.
Awesome!

STEP 6

Now, we **integrate** our expanded polynomial term by term.
Remember the power rule: the integral of tnt^n is tn+1n+1\frac{t^{n+1}}{n+1}.
Don't forget to add the constant of integration, which we'll call CC, but since it's a definite integral, the constant will disappear later.

STEP 7

∫(1−3t2+3t4−t6) dt=t−3t33+3t55−t77+C=t−t3+35t5−17t7+C\int (1 - 3t^2 + 3t^4 - t^6) \, dt = t - \frac{3t^3}{3} + \frac{3t^5}{5} - \frac{t^7}{7} + C = t - t^3 + \frac{3}{5}t^5 - \frac{1}{7}t^7 + C Look at that beautiful integral!

STEP 8

Time to **evaluate** our definite integral from 0 to -1.
Remember, this means we plug in the **upper limit** (−1-1) and subtract the result of plugging in the **lower limit** (00).

STEP 9

Let's plug in t=−1t = -1: (−1)−(−1)3+35(−1)5−17(−1)7=−1−(−1)+35(−1)−17(−1)=−1+1−35+17=−35+17(-1) - (-1)^3 + \frac{3}{5}(-1)^5 - \frac{1}{7}(-1)^7 = -1 - (-1) + \frac{3}{5}(-1) - \frac{1}{7}(-1) = -1 + 1 - \frac{3}{5} + \frac{1}{7} = -\frac{3}{5} + \frac{1}{7}

STEP 10

Now, let's plug in t=0t = 0: (0)−(0)3+35(0)5−17(0)7=0(0) - (0)^3 + \frac{3}{5}(0)^5 - \frac{1}{7}(0)^7 = 0

STEP 11

Finally, subtract the result for t=0t=0 from the result for t=−1t=-1: (−35+17)−0=−35+17=−2135+535=−1635 \left(-\frac{3}{5} + \frac{1}{7}\right) - 0 = -\frac{3}{5} + \frac{1}{7} = -\frac{21}{35} + \frac{5}{35} = -\frac{16}{35}

STEP 12

The definite integral ∫0−1(1−t2)3 dt\int_{0}^{-1}(1-t^{2})^{3} \, dt is equal to −1635-\frac{16}{35}.

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