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To solve the integral [tex]\(\int_{2}^{3} x^2 \sin x \, dx\)[/tex], we'll proceed with integration by parts. This method uses the formula:
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
Here, we need to identify parts of our function to assign to [tex]\(u\)[/tex] and [tex]\(dv\)[/tex]. Let's choose:
[tex]\[ u = x^2 \][/tex]
[tex]\[ dv = \sin x \, dx \][/tex]
We need to find [tex]\(du\)[/tex] and [tex]\(v\)[/tex]:
[tex]\[ du = \frac{d}{dx}(x^2) \, dx = 2x \, dx \][/tex]
[tex]\[ v = \int \sin x \, dx = -\cos x \][/tex]
Now, applying the integration by parts formula,
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
substitute the values of [tex]\(u\)[/tex], [tex]\(v\)[/tex], [tex]\(du\)[/tex], we get:
[tex]\[ \int_{2}^{3} x^2 \sin x \, dx = \left[ -x^2 \cos x \right]_{2}^{3} - \int_{2}^{3} -\cos x \cdot 2x \, dx \][/tex]
[tex]\[ = \left[ -x^2 \cos x \right]_{2}^{3} + 2 \int_{2}^{3} x \cos x \, dx \][/tex]
Next, we need to evaluate the second integral, [tex]\(\int_{2}^{3} x \cos x \, dx\)[/tex]. We'll again use integration by parts. Choose:
[tex]\[ u = x \][/tex]
[tex]\[ dv = \cos x \, dx \][/tex]
Then,
[tex]\[ du = dx \][/tex]
[tex]\[ v = \int \cos x \, dx = \sin x \][/tex]
Applying the formula again:
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
[tex]\[ \int_{2}^{3} x \cos x \, dx = \left[ x \sin x \right]_{2}^{3} - \int_{2}^{3} \sin x \, dx \][/tex]
[tex]\[ = \left[ x \sin x \right]_{2}^{3} - \left[ -\cos x \right]_{2}^{3} \][/tex]
[tex]\[ = \left[ x \sin x + \cos x \right]_{2}^{3} \][/tex]
Combining all parts together:
[tex]\[ \int_{2}^{3} x^2 \sin x \, dx = \left[ -x^2 \cos x \right]_{2}^{3} + 2 \left( \left[ x \sin x + \cos x \right]_{2}^{3} \right) \][/tex]
Evaluate at the bounds [tex]\(2\)[/tex] and [tex]\(3\)[/tex]:
[tex]\[ \left[ -3^2 \cos 3 - (-2^2 \cos 2) \right] + 2 \left( \left( 3 \sin 3 - 2 \sin 2 \right) + \left( \cos 3 - \cos 2 \right) \right) \][/tex]
[tex]\[ = \left[ -9 \cos 3 + 4 \cos 2 \right] + 2 \left( 3 \sin 3 - 2 \sin 2 + \cos 3 - \cos 2 \right) \][/tex]
[tex]\[ = -9 \cos 3 + 4 \cos 2 + 6 \sin 3 - 4 \sin 2 + 2 \cos 3 - 2 \cos 2 \][/tex]
[tex]\[ = -9 \cos 3 + 2 \cos 3 + 4 \cos 2 - 2 \cos 2 + 6 \sin 3 - 4 \sin 2 \][/tex]
[tex]\[ = -7 \cos 3 + 2 \cos 2 + 6 \sin 3 - 4 \sin 2 \][/tex]
Thus, the evaluated integral is:
[tex]\[ \int_{2}^{3} x^2 \sin x \, dx = -4 \sin 2 + 2 \cos 2 + 6 \sin 3 - 7 \cos 3 \][/tex]
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
Here, we need to identify parts of our function to assign to [tex]\(u\)[/tex] and [tex]\(dv\)[/tex]. Let's choose:
[tex]\[ u = x^2 \][/tex]
[tex]\[ dv = \sin x \, dx \][/tex]
We need to find [tex]\(du\)[/tex] and [tex]\(v\)[/tex]:
[tex]\[ du = \frac{d}{dx}(x^2) \, dx = 2x \, dx \][/tex]
[tex]\[ v = \int \sin x \, dx = -\cos x \][/tex]
Now, applying the integration by parts formula,
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
substitute the values of [tex]\(u\)[/tex], [tex]\(v\)[/tex], [tex]\(du\)[/tex], we get:
[tex]\[ \int_{2}^{3} x^2 \sin x \, dx = \left[ -x^2 \cos x \right]_{2}^{3} - \int_{2}^{3} -\cos x \cdot 2x \, dx \][/tex]
[tex]\[ = \left[ -x^2 \cos x \right]_{2}^{3} + 2 \int_{2}^{3} x \cos x \, dx \][/tex]
Next, we need to evaluate the second integral, [tex]\(\int_{2}^{3} x \cos x \, dx\)[/tex]. We'll again use integration by parts. Choose:
[tex]\[ u = x \][/tex]
[tex]\[ dv = \cos x \, dx \][/tex]
Then,
[tex]\[ du = dx \][/tex]
[tex]\[ v = \int \cos x \, dx = \sin x \][/tex]
Applying the formula again:
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
[tex]\[ \int_{2}^{3} x \cos x \, dx = \left[ x \sin x \right]_{2}^{3} - \int_{2}^{3} \sin x \, dx \][/tex]
[tex]\[ = \left[ x \sin x \right]_{2}^{3} - \left[ -\cos x \right]_{2}^{3} \][/tex]
[tex]\[ = \left[ x \sin x + \cos x \right]_{2}^{3} \][/tex]
Combining all parts together:
[tex]\[ \int_{2}^{3} x^2 \sin x \, dx = \left[ -x^2 \cos x \right]_{2}^{3} + 2 \left( \left[ x \sin x + \cos x \right]_{2}^{3} \right) \][/tex]
Evaluate at the bounds [tex]\(2\)[/tex] and [tex]\(3\)[/tex]:
[tex]\[ \left[ -3^2 \cos 3 - (-2^2 \cos 2) \right] + 2 \left( \left( 3 \sin 3 - 2 \sin 2 \right) + \left( \cos 3 - \cos 2 \right) \right) \][/tex]
[tex]\[ = \left[ -9 \cos 3 + 4 \cos 2 \right] + 2 \left( 3 \sin 3 - 2 \sin 2 + \cos 3 - \cos 2 \right) \][/tex]
[tex]\[ = -9 \cos 3 + 4 \cos 2 + 6 \sin 3 - 4 \sin 2 + 2 \cos 3 - 2 \cos 2 \][/tex]
[tex]\[ = -9 \cos 3 + 2 \cos 3 + 4 \cos 2 - 2 \cos 2 + 6 \sin 3 - 4 \sin 2 \][/tex]
[tex]\[ = -7 \cos 3 + 2 \cos 2 + 6 \sin 3 - 4 \sin 2 \][/tex]
Thus, the evaluated integral is:
[tex]\[ \int_{2}^{3} x^2 \sin x \, dx = -4 \sin 2 + 2 \cos 2 + 6 \sin 3 - 7 \cos 3 \][/tex]
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